Question 13 Marks
$113 \times 87$
AnswerHere, we will use the identity $(a-b)(a+b)=a^2-b^2$
Let us consider the following expression:
$113 \times 87$
$\because \frac{113+87}{2}=\frac{200}{2}=100$, therefore, we will write the above product as:
$113 \times 87$
$=(100+13)(100-13)$
$=(100)^2-(13)^2$
$=10000-169$
$=9831$
Thus, the answer is 9831
View full question & answer→Question 23 Marks
Find the product:
$1.5 x\left(10 x^2 y-100 x y^2\right)$
AnswerTo find the product, we will use distributive law as follows:
$1.5 x\left(10 x^2 y-100 x y^2\right) \\
=\left(1.5 x \times 10 x^2 y\right)-\left(1.5 x \times 100 x y^2\right) \\
=\left(15 x^{1+2} y\right)-\left(150 x^{1+1} y^2\right) \\
=15 x^3 y-150 x^2 y^2$
Thus, the answer is $15 x^3 y-150 x^2 y^2$
View full question & answer→Question 33 Marks
Simplify following:
$\text{x}^2 − 3\text{x} + 5 − \frac{1}{2} (3\text{x}^2 − 5\text{x} + 7)$
Answer$\text{x}^2 − 3\text{x} + 5 − \frac{1}{2} (3\text{x}^2 − 5\text{x} + 7)$
$=\text{x}^2 − 3\text{x} + 5 − \frac{3\text{x}^2}{2} +\frac{5\text{x}}{2}-\frac{ 7}{2}$
$=\text{x}^2 − \frac{3\text{x}^2}{2}- 3\text{x} +\frac{5\text{x}}{2}+ 5 -\frac{ 7}{2}$ (Collecting like terms)
$=\big(\frac{1−3}{2})\text{x}^2+\big(\frac{−3+5}{2}\big)\text{x}+\big(\frac{10−7}{2}\big)$
$=−\frac{\text{x}^2}{2}−\frac{\text{x}}{2}+\frac{3}{2}$
Thus, the answer is $=−\frac{\text{x}^2}{2}−\frac{\text{x}}{2}+\frac{3}{2}$
View full question & answer→Question 43 Marks
Find the value of x, if:
$4 x=(52)^2-(48)^2$
AnswerLet us consider the following equation:
$4 x=(52)^2-(48)^2$
Using the identity $(a+b)(a-b)=a^2-b^2$ we get:
$4 x=(52)^2-(48)^2$
$4 x=(52+48)(52-48)$
$4 x=100 \times 4=400$
$\Rightarrow 4 x=400$
$\Rightarrow x=100 \text { (Dividing both sides by } 4 \text { ) }$
View full question & answer→Question 53 Marks
Simplify following:
$−\frac{1}{2}\text{a}^2\text{b}^2\text{c}+\frac{1}{3}\text{ab}^2\text{c}−\frac{1}{4}\text{abc}^2−\frac{1}{5}\text{cb}^2\text{a}^2\\+\frac{1}{6}\text{cb}^2\text{a}−\frac{1}{7}\text{c}^2\text{ab}+\frac{1}{8}\text{ca}^2\text{b}.$
Answer$−\frac{1}{2}\text{a}^2\text{b}^2\text{c}+\frac{1}{3}\text{ab}^2\text{c}−\frac{1}{4}\text{abc}^2−\frac{1}{5}\text{cb}^2\text{a}^2\\+\frac{1}{6}\text{cb}^2\text{a}−\frac{1}{7}\text{c}^2\text{ab}+\frac{1}{8}\text{ca}^2\text{b}.$
$−\frac{1}{2}\text{a}^2\text{b}^2\text{c}-\frac{1}{5}\text{cb}^2\text{a}^2+\frac{1}{3}\text{ab}^2\text{c}+\frac{1}{6}\text{cb}^2\text{a}−\frac{1}{4}\text{abc}^2\\-\frac{1}{7}\text{c}^2\text{ab}+\frac{1}{8}\text{ca}^2\text{b}.$
$=\big(\frac{−5−2}{10}\big)\text{a}^2\text{b}^2\text{c} +\big(\frac{2+1}{6}\big)\text{ab}^2\text{a}^2 \\+\big(\frac{−7−4}{28}\big)\text{c}^2\text{ab}+\frac{1}{8}\text{ca}^2\text{b}$ (Collecting like terms)
$=−\frac{7}{10}\text{a}^2\text{b}^2\text{c}+\frac{1}{2}\text{ab}^2\text{c}−\frac{11}{28}\text{abc}^2+\frac{1}{8}\text{a}^2\text{bc} $ (Combining like terms)
View full question & answer→Question 63 Marks
Express product as a monomials and verify the result for x = 1, y = 2:
$\big(−\frac{4}{7}\text{a}^2\text{b}\big)×\big(−\frac{2}{3}\text{b}^2\text{c}\big)×\big(−\frac{7}{6}\text{c}^2\text{a}\big)$
AnswerTo multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a ^{ m } \times a ^{ n }= a ^{ m + n }$. We have: $\left(-\frac{4}{7} a ^2 b\right) \times\left(-\frac{2}{3} b^2 c \right) \times\left(-\frac{7}{6} c ^2 a \right)$
$=\big[\big(−\frac{4}{7}\big)×\big(−\frac{2}{3}\big)×\big(−\frac{7}{6}\big)\big]×\big(\text{a}^2×\text{a}\big)\\×\big(\text{b}×\text{b}^2\big)×\big(\text{c}×\text{c}^2)$
$=\big[\big(−\frac{4}{7}\big)×\big(−\frac{2}{3}\big)×\big(−\frac{7}{6}\big)\big]×\big(\text{a}^{2+1}\big)\\×\big(\text{b}^{1+2}\big)×\big(\text{c}^{1+2}\big)$
$=−\frac{4}{9}\text{a}^3\text{b}^3\text{c}^3$
$\because$ The expression doesn't consist of the variables x and y.
$\therefore$ The result cannot be verified for x = 1 and y = 2. Thus, the answer is $-\frac{4}{9}\text{a}^3\text{b}^3\text{c}^3$
View full question & answer→Question 73 Marks
If $x^2+y^2=29$ and $x y=2$, find the value of
$x+y$
AnswerWe have:
$(\text{x}+\text{y})^2=\text{x}^2+2\text{xy}+\text{y}^2$
$⇒(\text{x}+{\text{y}})=±\sqrt{\text{x}^2+2\text{xy}+\text{y}^2}$
$⇒(\text{x}+{\text{y}})=±\sqrt{29+2\times2}$ $\big( ∵\text{ x}^2+\text{y}^2=29 \ \text{and} \text{ xy}=2\big)$
$⇒(\text{x}+{\text{y}})=±\sqrt{29+4}$
$⇒(\text{x}+{\text{y}})=±\sqrt{33}$
View full question & answer→Question 83 Marks
Find product:
$\big(\frac{2}{5}\text{a}^2\text{b}\big)×(−15\text{b}^2\text{ac}\big)×\big(\frac{−1}{2}\text{c}^2\big)$
AnswerTo multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a ^{ m } \times a ^{ n }= a ^{ m + n }$.
We have:
$\left(\frac{2}{5} a ^2 b \right) \times\left(-15 b ^2 a c \right) \times\left(\frac{-1}{2} c ^2\right)=\left[-\frac{2}{5} \times(-15) \times\left(-\frac{1}{2}\right)\right] \times\left( a ^2 \times a \right)$
$=\big[-\frac{2}{5}×(−15)×(−\frac{1}{2})\big]×\big(\text{a}^{2+1}\big)\\×\big(\text{b}^{1+2}\big)×\big(\text{c}^{1+2}\big)$
$=3\text{a}^3\text{b}^3\text{c}^3$
$\because$ The expression doesn't consist of the variables x and y.
$\therefore$ The result cannot be verified for x = 1 and y = 2
Thus, the answer is $3\text{a}^3\text{b}^3\text{c}^3$
View full question & answer→Question 93 Marks
If $3 x+5 y=11$ and $x y=2$, find the value of $9 x^2+25 y^2$
AnswerWe have:
$(3 x+5 y)^2=(3 x)^2+2(3 x)(5 y)+(5 y)^2$
$\Rightarrow(3 x+5 y)^2=9 x^2+30 x y+25 y^2$
$\Rightarrow 9 x^2+25 y^2=(3 x+5 y)^2-30 x y$
$\Rightarrow 9 x^2+25 y^2=112-30 \times 2(\because 3 x+5 y=11 \text { and } x y=2)$
$\Rightarrow 9 x^2+25 y^2=121-60$
$\Rightarrow 9 x^2+25 y^2=61$
View full question & answer→Question 103 Marks
Find the product:
$\frac{4}{3}\text{a}\big(\text{a}^2 +\text{ b}^2 − 3\text{c}^2\big)$
AnswerTo find the product, we will use distributive law as follows:$\frac{4}{3}\text{a}\big(\text{a}^2 +\text{ b}^2 − 3\text{c}^2\big)$
$=\frac{4}{3}\text{a}×\text{a}^2+\frac{4}{3}\text{a}×\text{b}^2−\frac{4}{3}\text{a}×3\text{c}^2$
$=\frac{4}{3}\text{a}^{1+2}+\frac{4}{3}\text{ab}^2−4\text{ac}^2$
$=\frac{4}{3}\text{a}^3+\frac{4}{3}\text{ab}^2−4\text{ac}^2$
Thus, the answer is $\frac{4}{3}\text{a}^3+\frac{4}{3}\text{ab}^2−4\text{ac}^2$
View full question & answer→Question 113 Marks
Multiply:
$\big(\frac{3}{5}\text{x}+\frac{1}{2}\text{y}\big) \ \text{by} \ \big(\frac{5}{6}\text{x}+4\text{y}\big)$
AnswerTo multiply, we will use distributive law as follows:
$\big(\frac{3}{5}\text{x}+\frac{1}{2}\text{y}\big) \ \text{by} \ \big(\frac{5}{6}\text{x}+4\text{y}\big)$
$=\frac{3}{5}\text{x}\big(\frac{5}{6}\text{x}+4\text{y}\big)+\frac{1}{2}\text{y}\big(\frac{5}{6}\text{x}+4\text{y})$
$=\frac{1}{2}\text{x}^2+\frac{12}{5}\text{xy}+\frac{5}{12}\text{xy}+2\text{y}^2$
$=\frac{1}{2}\text{x}^2+\big(\frac{144+25}{60}\big)\text{xy}+2\text{y}^2$
$=\frac{1}{2}\text{x}^2+\frac{169}{60}\text{xy}+2\text{y}^2$
Thus, the answer is $=\frac{1}{2}\text{x}^2+\frac{169}{60}\text{xy}+2\text{y}^2.$
View full question & answer→Question 123 Marks
Simplify:
$2 x^2\left(x^3-x\right)-3 x\left(x^4+2 x\right)-2\left(x^4-3 x^2\right)$
AnswerTo simplify, we will use distributive law as follows:
$2 x^2\left(x^3-x\right)-3 x\left(x^4+2 x\right)-2\left(x^4-3 x^2\right) \\
=2 x^5-2 x^3-3 x^5-6 x^2-2 x^4+6 x^2 \\
=2 x^5-3 x^5-2 x^4-2 x^3-6 x^2+6 x^2 \\
=-x^5-2 x^4-2 x^3$
View full question & answer→Question 133 Marks
Simplify:
$2 a^2+3 a\left(1-2 a^3\right)+a(a+1)$
AnswerTo simplify, we will use distributive law as follows:
$2 a^2+3 a\left(1-2 a^3\right)+a(a+1)$
$=2 a^2+3 a-6 a^4+a^2+a$
$=2 a^2+a^2+3 a+a-6 a^4$
$=3 a^2+4 a-6 a^4$
View full question & answer→Question 143 Marks
Find the product:
$-11 y^2(3 y+7)$
AnswerTo find the product, we will use distributive law as follows:
$-11 y^2(3 y+7)$
$=\left(-11 y^2\right) \times 3 y+\left(-11 y^2\right) \times 7$
$=(-11 \times 3)\left(y^2 \times y\right)+(-11 \times 7) \times\left(y^2\right)$
$=(-33)\left(y^2+1\right)+(-77) \times\left(y^2\right)$
$=-33 y^3-77 y^2$
Thus, the answer is $-33 y^3-77 y^2$.
View full question & answer→Question 153 Marks
Find the product:
$250.5\text{xy}\big(\text{xz}+\frac{\text{y}}{10}\big)$
AnswerTo find the product, we will use distributive law as follows:$250.5\text{xy}\big(\text{xz}+\frac{\text{y}}{10}\big)$
$=250.5\text{xy}×\text{xz}+250.5\text{xy}×\frac{\text{y}}{10}$
$=250.5\text{x}^{1+1}\text{yz}+25.05\text{xy}^{1+1}$
$=250.5\text{x}^2\text{yz}+25.05\text{xy}^2$
Thus, the answer is $250.5\text{x}^2\text{yz}+25.05\text{xy}^2$
View full question & answer→Question 163 Marks
Take away:
$\frac{5\text{a}^2}{2}+\frac{3\text{a}^3}{2}+\frac{\text{a}}{3}−\frac{6}{5}$ from $\frac{1}{3}\text{a}^3−\frac{3}{4}\text{a}^2−\frac{5}{2}$
AnswerThe difference is given by:$\big(\frac{1}{3}\text{a}^3−\frac{3}{4}\text{a}^2−\frac{5}{2}\big)-\\\big(\frac{5\text{a}^2}{2}+\frac{3\text{a}^3}{2}+\frac{\text{a}}{3}−\frac{6}{5}\big)$
$=\frac{1}{3}\text{a}^3−\frac{3}{4}\text{a}^2−\frac{5}{2}-\frac{5\text{a}^2}{2}\\-\frac{3\text{a}^3}{2}-\frac{\text{a}}{3}+\frac{6}{5}$
$=\frac{1}{3}\text{a}^3−\frac{3\text{a}^{3}}2−\frac{3\text{a}^2}{4}-\frac{5\text{a}^{2}}{2}−\frac{\text{a}}{3}\\−\frac{5}{2}+\frac{6}{5} $
$=\big(\frac{2−9}{6}\big)\text{a}^3+\big(\frac{−3−10}{4})\text{a}^2\\−\frac{\text{a}}{3}+\big(\frac{−25+12}{10}\big)$ (Collecting like terms)
$=-\frac{7}{6}\text{a}^3-\frac{13}{4}\text{a}^2-\frac{\text{a}}{3}-\frac{13}{10 } $ (Combining like terms)
View full question & answer→Question 173 Marks
Find the product:
$4.1 x y(1.1 x-y)$
AnswerTo find the product, we will use distributive law as follows:
$4.1 x y(1.1 x-y)$
$=(4.1 x y \times 1.1 x)-(4.1 x y \times y)$
$=[(4.1 \times 1.1) \times x y \times x]-(4.1 x y \times y)$
$=\left(4.51 x^{1+1} y\right)-\left(4.1 x y^{1+1}\right)$
$=4.51 x^2 y-4.1 x y^2$
Thus, the answer is $4.51 x^2 y-4.1 x y^2$
View full question & answer→Question 183 Marks
Find the product:
$−\frac{4}{27}\text{xyz}\big(\frac{9}{2}\text{x}^2\text{yz}−\frac{3}{4}\text{xyz}^2\big)$
AnswerTo find the product, we will use distributive law as follows:
$−\frac{4}{27}\text{xyz}\big(\frac{9}{2}\text{x}^2\text{yz}−\frac{3}{4}\text{xyz}^2\big)$
$=\Big\{\big(-\frac{4}{27}\text{xyz}\big)\big(\frac{9}{2}\text{x}^2\text{yz}\big)\Big\}\\−\Big\{(-\frac{4}{27}\text{xyz}\big)\big(\frac{3}{4}\text{xyz}^2\big)\Big\}$
$=\Big\{\big(−\frac{4}{27}×\frac{9}{2}\big)\big(\text{x}^{1+2}\text{y}^{1+1}\text{z}^{1+1}\big)\Big\}\\−\Big\{\big(−\frac{4}{27}×\frac{3}{4}\big)\big(\text{x}^{1+1}\text{y}^{1+1}\text{z}^{1+2}\big)\Big\}$
$=−\frac{2}{3}\text{x}^3\text{y}^2\text{z}^2+\frac{1}{9}\text{x}^2\text{y}^2\text{z}^3$
Thus, the answer is $−\frac{2}{3}\text{x}^3\text{y}^2\text{z}^2+\frac{1}{9}\text{x}^2\text{y}^2\text{z}^3$
View full question & answer→Question 193 Marks
Simplify:
$4 a b(a-b)-6 a^2\left(b-b^2\right)-3 b^2\left(2 a^2-a\right)+2 a b(b-a)$
AnswerTo simplify, we will use distributive law as follows:
$4 a b(a-b)-6 a^2\left(b-b^2\right)-3 b^2\left(2 a^2-a\right)+2 a b(b-a)$
$=4 a^2 b-4 a b^2-6 a^2 b+6 a^2 b^2-6 b^2 a^2+3 b^2 a+2 a b^2-2 a^2 b$
$=4 a^2 b-6 a^2 b-2 a^2 b-4 a b^2+3 b^2 a+2 a b^2+6 a^2 b^2-6 b^2 a^2$
$=-4 a^2 b+a b^2$
View full question & answer→Question 203 Marks
Add the following algebric expressions:
$\frac{11}{2}\text{xy}+\frac{12}{5}\text{y}+\frac{13}{7}\text{x},$$-\frac{11}{2}\text{y}-\frac{12}{5}\text{x}-\frac{13}{7}\text{xy}$
AnswerTo add we proceed as follows:
$\big(\frac{11}{2}\text{xy}-\frac{12}{5}\text{y}+\frac{13}{7}\text{x}\big)\\+\big(-\frac{11}{2}\text{y}-\frac{12}{5}\text{x}-\frac{13}{7}\text{xy}\big)$
$=\frac{11}{2}\text{xy}+\frac{12}{5}\text{y}+\frac{13}{7}\text{x}-\frac{11}{2}\text{y}\\-\frac{12}{5}\text{x}-\frac{13}{7}\text{xy}$
$=\frac{11}{2}\text{xy}-\frac{13}{7}\text{xy}+\frac{12}{5}\text{y}-\frac{11}{2}\text{y}\\+\frac{13}{7}\text{x}-\frac{12}{5}\text{x}$ (Combining like terms)
$=\frac{51}{14}\text{xy}-\frac{31}{10}\text{y}-\frac{19}{35}\text{x}$ (Combining like terms)
View full question & answer→Question 213 Marks
If $x+y=4$ and $x y=2$, find the value of $x^2+y^2$
AnswerWe have:
$(x+y)^2=x^2+2 x y+y^2$
$\Rightarrow x^2+y^2=(x+y)^2-2 x y+y^2$
$\Rightarrow x^2+y^2=4^2-2 \times 2(\because x+y=4 \text { and } x y=2)$
$\Rightarrow x^2+y^2=16-4$
$\Rightarrow x^2+y^2=12$
View full question & answer→Question 223 Marks
$1.8 \times 2.2$
AnswerHere, we will use the identity $(a-b)(a+b)=a^2-b^2$
Let us consider the following expression:
$1.8 \times 2.2$
$\because \frac{1.8+2.2}{2}=\frac{4}{2}=2$, therefore, we will write the above product as:
$1.8 \times 2.2$
$=(2-0.2)(2+0.2)$
$=(2)^2-(0.2)^2$
$=4-0.04$
$=3.96$
Thus, the answer is 3.96
View full question & answer→Question 233 Marks
Multiply:
$(7 x+y) \text { by }(x+5 y)$
AnswerTo multiply, we will use distributive law as follows:
$(7 x+y) \text { by }(x+5 y)$
$=7 x(x+5 y)+y(x+5 y)$
$=7 x^2+35 x y+x y+5 y^2$
$=7 x^2+36 x y+5 y^2$
Thus, the answer is $7 x^2+36 x y+5 y^2$.
View full question & answer→Question 243 Marks
Find the value of $x$, if:
$14 x=(47)^2-(33)^2$
AnswerLet us consider the following equation:
$14 x=(47)^2-(33)^2$
Using the identity $(a+b)(a-b)=a^2-b^2$
we get:
$14 x=(47)^2-(33)^2$
$14 x=(47+33)(47-33)$
$14 x=80 \times 14=1120$
$\Rightarrow 14 x=1120$
$\Rightarrow x=80 \text { (Dividing both sides by } 14 \text { ) }$
View full question & answer→Question 253 Marks
If $x^2+y^2=29$ and $x y=2$, find the value of
$x-y$
AnswerWe have:
$(\text{x}-\text{y})^2=\text{x}^2-2\text{xy}+\text{y}^2$
$⇒(\text{x}-{\text{y}})=±\sqrt{\text{x}^2-2\text{xy}+\text{y}^2}$
$⇒(\text{x}+{\text{y}})=±\sqrt{29-2\times2}$ $\big( ∵\text{ x}^2+\text{y}^2=29 \ \text{and} \text{ xy}=2\big)$
$⇒(\text{x}+{\text{y}})=±\sqrt{29-4}$
$⇒(\text{x}+{\text{y}})=±\sqrt{25}$
$⇒(\text{x}+{\text{y}})=±5$
View full question & answer→Question 263 Marks
$197 \times 203$
AnswerHere, we will use the identity $(a-b)(a+b)=a^2-b^2$
Let us consider the following expression:
$197 \times 203$
$\because \frac{197+203}{2}=\frac{400}{2}=200$, therefore, we will write the above product as:
$197 \times 203$
$=(200-3)(200+3)$
$=(200)^2-(3)^2$
$=40000-9$
$=39991$
Thus, the answer is 39991
View full question & answer→Question 273 Marks
Find the product:
$0.1 y\left(0.1 x^5+0.1 y\right)$
AnswerTo find the product, we will use distributive law as follows:
$0.1 y\left(0.1 x^5+0.1 y\right) \\
=(0.1 y)\left(0.1 x^5\right)+(0.1 y)(0.1 y) \\
=(0.1 \times 0.1)\left(y \times x^5\right)+(0.1 \times 0.1)(y \times y) \\
=(0.1 \times 0.1)\left(x^5 \times y\right)+(0.1 \times 0.1)\left(y^{1+1}\right) \\
=0.01 x^5 y+0.01 y^2$
Thus, the answer is $0.01 x^5 y+0.01 y^2$
View full question & answer→Question 283 Marks
Express product as a monomials and verify the result for x = 1, y = 2:
$\big(\frac{4}{9}\text{abc}^3\big)×\big(−\frac{27}{5}\text{a}^3\text{b}^2)×\big(−8\text{b}^3\text{c}\big)$
AnswerTo multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a^m \times a^n=a^{m+n}$.
We have:
$\big(\frac{4}{9}\text{abc}^3\big)×\big(−\frac{27}{5}\text{a}^3\text{b}^2)×\big(−8\text{b}^3\text{c}\big)$
$\big[\big(\frac{4}{9}\big)×\big(−\frac{27}{5}\big)×(−8)\big]×\big(\text{a}×\text{a}^3\big)\\×\big(\text{b}×\text{b}^2×\text{b}^3\big)×\big(\text{c}^3×\text{c}\big)$
$\big[\big(\frac{4}{9}\big)×\big(−\frac{27}{5}\big)×(−8)\big]×\big(\text{a}^{1+3}\big)\\×\big(\text{b}^{1+2+3}\big)×\big(\text{c}^{3+1}\big)$
$=\frac{96}{5}\text{a}^4\text{b}^6\text{c}^4.$ Thus, the answer is $\frac{96}{5}\text{a}^4\text{b}^6\text{c}^4.$
$\because$ The expression doesn't consist of the variables x and y.
$\therefore$ The result cannot be verified for x = 1 and y = 2
View full question & answer→Question 293 Marks
Find product:
$(0.5\text{x})×\big(\frac{1}{3}\text{xy}^2\text{z}^4)×\big(24\text{x}^2\text{yz}\big)$
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^m \times a^n=a^{m+n}$.
We have:
$(0.5\text{x})×\big(\frac{1}{3}\text{xy}^2\text{z}^4)×\big(24\text{x}^2\text{yz}\big)$
$=\big(0.5×\frac{1}{3}×24)×\big(\text{x}×\text{x}×\text{x}^2)\\×\big(\text{y}^2×\text{y}\big)×\big(\text{z}^4×\text{z}\big)$
$=\big(0.5×\frac{1}{3}×24)×\big(\text{x}^{1+1+2}\big)\\×\big(\text{y}^{2+1}\big)×\big(\text{z}^{4+1}\big)$
$=4\text{x}^4\text{y}^3\text{z}^5$
Thus, the answer is $=4\text{x}^4\text{y}^3\text{z}^5$
View full question & answer→Question 303 Marks
Simplify following:
$\big(\frac{1}{3}\text{y}^2−\frac{4}{7}\text{y}+11\big)−(\frac{1}{7}\text{y}−3+2\text{y}^2\big)−\\\big(\frac{2}{7}\text{y}−\frac{2}{3}\text{y}^2+2\big)$
Answer$\big(\frac{1}{3}\text{y}^2−\frac{4}{7}\text{y}+11\big)−(\frac{1}{7}\text{y}−3+2\text{y}^2\big)−\\\big(\frac{2}{7}\text{y}−\frac{2}{3}\text{y}^2+2\big)$
$=\frac{1}{3}\text{y}^2−\frac{4}{7}\text{y}+11-\frac{1}{7}\text{y}+3-2\text{y}^2\\−\frac{2}{7}\text{y}+\frac{2}{3}\text{y}^2-2\big)$
$=\frac{1}{3}\text{y}^2+\frac{2}{3}\text{y}^2−2\text{y}^2-\frac{4}{7}\text{y}-\frac{1}{7}\text{y}\\-\frac{2}{7}\text{y}+11+3-2$ (Collecting like terms)
$=\big(\frac{1−6+2}{3}\big)\text{y}^2+\big(\frac{−4−1−2}{7}\big)\text{y}+12$(Combining like terms)
$= −\text{y}^2−7\text{y}+12$
View full question & answer→Question 313 Marks
What must be added to following expressions to make it a whole square?
$4 x^2-12 x+7$
AnswerLet us consider the following expression:
$4 x^2-12 x+7$
The above expression can be written as:
$4 x^2-12 x+7=(2 x)^2-2 \times 2 x \times 3+7$
It is evident that if $2 x$ is considered as the first term and 3 is considered as the second term, 2 is required to be added to the above expression to make it a perfect square.
Therefore, 7 must become 9 .
Therefore, adding and subtracting 2 in the above expression, we get:
$\left(4 x^2-12 x+7\right)+2-2$
$=\left\{(2 x)^2-2 \times 2 x \times 3+7\right\}+2-2$
$=\left\{(2 x)^2-2 \times 2 x \times 3+9\right\}-2$
$=(2 x+3)^2-2$
Thus, the answer is 2 .
View full question & answer→Question 323 Marks
Multiply the monomial by the binomial and find the value of $x=-1, y=0.25$ and $z=0.05$ :
$z^2(x-y)$
AnswerTo find the product, we will use distributive law as follows:
$z^2(x-y)$
$=z^2 \times x-z^2 \times y$
$=x z^2-y z^2$
Substituting $x=-1, y=0.25$ and $z=0.05$ in the result, we get:
$x z^2-y z^2$
$=(-1)(0.05)^2-(0.25)(0.05)^2$
$=(-1)(0.0025)-(0.25)(0.0025)$
$=-0.0025-0.000625$
$=-0.003125$
View full question & answer→Question 333 Marks
Find the value of x, if:
$5 x=(50)^2-(40)^2$
AnswerLet us consider the following equation:
$5 x=(50)^2-(40)^2$
Using the identity $(a+b)(a-b)=a^2-b^2$
we get:
$5 x=(50)^2-(40)^2$
$5 x=(50+40)(50-40)$
$5 x=90 \times 10=900$
$\Rightarrow 5 x=900$
$\Rightarrow x=180 \text { (Dividing both sides by } 5)$
View full question & answer→Question 343 Marks
Subtract:
$\frac{2}{3}\text{y}^{3}+\frac{2}{7}\text{y}^{2}-5$ from $\frac{1}{3}\text{y}^{3}+\frac{5}{7}\text{y}^{2}+\text{y}-2$
Answer$\big(\frac{1}{3}\text{y}^{3}+\frac{5}{7}\text{y}^{2}+\text{y}-2\big)-\big(\frac{2}{3}\text{y}^{3}-\frac{2}{7}\text{y}^{2}-5\big)$
$=\frac{1}{3}\text{y}^{3}+\frac{5}{7}\text{y}^{2}+\text{y}-2-\frac{2}{3}\text{y}^{3}+\frac{2}{7}\text{y}^{2}+5$
$=\frac{1}{3}\text{y}^{3}-\frac{2}{3}\text{y}^{3}+\frac{5}{7}\text{y}^{2}+\frac{2}{7}\text{y}^{2}+\text{y}-2+5$ (Collecting like terms)
$=-\frac{1}{3}\text{y}^{3}+\text{y}^{2}+\text{y}+3$ (Combining like terms)
View full question & answer→Question 353 Marks
Simplify:
$(x-y)(x+y)\left(x^2+y^2\right)\left(x^4+y^2\right)$
AnswerTo simplify, we will proceed as follows:
$(x-y)(x+y)\left(x^2+y^2\right)\left(x^4+y^4\right)$
$=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left[\because(a+b)(a-b)=a^2-b^2\right]$
$=\left(x^4-y^4\right)\left(x^4+y^4\right)$
$=x^8-x^8$
View full question & answer→Question 363 Marks
Simplify:
$a^2 b^2(a+2 b)(3 a+b)$
AnswerTo simplify, we will proceed as follows: $a^2 b^2(a+2 b)(3 a+b)$
$=\left[a^2 b^2(a+2 b)\right](3 a+b)$
$=\left(a^3 b^2+2 a^2 b^3(3 a+b)\right.$
$=3 a\left(a^3 b^2+2 a^2 b^3\right)+b\left(a^3 b^2+2 a^2 b^3\right)$
$=3 a^4 b^2+6 a^3 b^3+a^3 b^3+2 a^2 b^4 .$
$=3 a^4 b^2+7 a^3 b^3+2 a^2 b^4$
Thus, the answer is $3 a^4 b^2+7 a^3 b^3+2 a^2 b^4$
View full question & answer→Question 373 Marks
Simplify:
$x^2(x+2 y)(x-3 y)$
AnswerTo simplify, we will proceed as follows:
$x^2(x+2 y)(x-3 y)$
$=\left[x^2(x+2 y)\right](x-3 y)$
$=\left(x^3+2 x^2 y\right)(x-3 y)$
$=x^3(x-3 y)+2 x^2 y(x-3 y)$
$=x^4-3 x^3 y+2 x^3 y-6 x^2 y^2$
$=x^4-x^3 y-6 x^2 y^2$
Thus, the answer is $x^4-x^3 y-6 x^2 y^2$.
View full question & answer→Question 383 Marks
Multiply:
$(5 x+3) \text { by }(7 x+2)$
AnswerTo multiply, we will use distributive law as follows:
$(5 x+3)(7 x+2)$
$=5 x(7 x+2)+3(7 x+2)$
$=(5 x \times 7 x+5 x \times 2)+(3 \times 7 x+3 \times 2)$
$=\left(35 x^2+10 x\right)(21 x+6)$
$=35 x^2+10 x+21 x+6$
$=35 x^2+31 x+6$
Thus, the answer is $35 x^2+31 x+6$.
View full question & answer→Question 393 Marks
Take away:
$\frac{7}{4}\text{x}^3+\frac{3}{5}\text{x}^2+\frac{1}{2}\text{x}+\frac{9}{2}$ from $\frac{7}{2}-\frac{\text{x}}{3}-\frac{\text{x}^2}{5}$
AnswerHe difference is given by:
$\big(\frac{7}{2}-\frac{\text{x}}{3}-\frac{\text{x}^2}{5}\big)-\big(\frac{7\text{x}^3}{4}+\frac{3\text{x}^2}{5}+\frac{\text{x}}{2}+\frac{9}{2}\big)$
$=\frac{7}{2}-\frac{\text{x}}{3}-\frac{\text{x}^2}{5}-\frac{7\text{x}^3}{4}-\frac{3\text{x}^2}{5}-\frac{\text{x}}{2}-\frac{9}{2}$
$=\frac{7}{2}-\frac{\text{9}}{2}-\frac{\text{x}}{3}-\frac{\text{x}}{2}-\frac{\text{x}^2}{5}-\frac{3\text{x}^2}{5}-\frac{7\text{x}^3}{4}$
$=\big(\frac{7-9}{2}\big)+\big(\frac{-2-3}{6})\text{x}+\big(\frac{-1-3}{5}\big)\text{x}^2-\frac{7\text{x}^3}{4}$ (Collecting like terms)
$=−1−\frac{5\text{x}}{6}−\frac{4\text{x}^2}{5}−\frac{7\text{x}^3}{4}$ (Combining like terms)
View full question & answer→Question 403 Marks
Multiply:
(0.8a - 0.5b) by (1.5a - 3b)
AnswerTo multiply, we will use distributive law as follows:
$(0.8 a-0.5 b)(1.5 a-3 b)$
$=0.8 a(1.5 a-3 b-0.5 b(1.5 a-3 b)$
$=2 a^2-2.4 a b-0.75 a b+1.5 b^2$
$=1.2 a^2-3.15 a b+1.5 b^2$
Thus, the answer is $1.2 a ^2-3.15 ab +1.5 b^2$.
View full question & answer→Question 413 Marks
Multiply:
$\big(\frac{\text{x}}{7}+\frac{\text{x}^2}{2}) \ \text{by}\big(\frac{2}{5}+\frac{9\text{x}}{4}\big)$
AnswerTo multiply, we will use distributive law as follows:$\big(\frac{\text{x}}{7}+\frac{\text{x}^2}{2}) \ \text{by}\big(\frac{2}{5}+\frac{9\text{x}}{4}\big)$
$=\frac{\text{x}}{7}\big(\frac{2}{5}+\frac{9\text{x}}{4}\big)+\frac{\text{x}^2}{2}\big(\frac{2}{5}+\frac{9\text{x}}{4}\big)$
$=\frac{2\text{x}}{35}+\frac{9\text{x}^2}{28}+\frac{\text{x}^2}{5}+\frac{9\text{x}^3}{8}$
$=\frac{2\text{x}}{35}+\big(\frac{45+28}{140}\big)\text{x}^2+ \frac{9\text{x}^3}{8}$
$=\frac{2x}{35}+\frac{73\text{x}^2}{140} +\frac{9\text{x}^2}{8}$
Thus, the answer is $\frac{2x}{35}+\frac{73\text{x}^2}{140} +\frac{9\text{x}^2}{8}.$
View full question & answer→Question 423 Marks
Simplify:
$\frac{3}{2}\text{x}^2\big(\text{x}^2−1\big)+\frac{1}{4}\text{x}^2\big(\text{x}^2+\text{x}\big)−\frac{3}{4}\text{x}\big(\text{x}^3−1\big)$
AnswerTo simplify, we will use distributive law as follows: $\frac{3}{2}\text{x}^2\big(\text{x}^2−1\big)+\frac{1}{4}\text{x}^2\big(\text{x}^2+\text{x}\big)−\frac{3}{4}\text{x}\big(\text{x}^3−1\big)$$=\frac{3}{2}\text{x}^4−\frac{3}{2}\text{x}^2+\frac{1}{4}\text{x}^4+\frac{1}{4}\text{x}^3−\frac{3}{4}\text{x}^4+\frac{3}{4}\text{x}$
$=\frac{3}{2}\text{x}^4+\frac{1}{4}\text{x}^4−\frac{3}{4}\text{x}^4+\frac{1}{4}\text{x}^3−\frac{3}{2}\text{x}^2+\frac{3}{4}\text{x}$
$=\big(\frac{6+1−3}{4}\big)\text{x}^4+\frac{1}{4}\text{x}^3−\frac{3}{2}\text{x}^2+\frac{3}{4}\text{x}$
$=\text{x}^4+\frac{1}{4}\text{x}^3−\frac{3}{2}\text{x}^2+\frac{3}{4}\text{x}$
View full question & answer→Question 433 Marks
Add the following algebric expressions:
$\frac{7}{2}\text{x}^{3}-\frac{1}{2}\text{x}^{2}+\frac{5}{3},$$\frac{3}{2}\text{x}^{3}+\frac{7}{4}\text{x}^{2}-\text{x}+\frac{1}{3},$$\frac{3}{2}\text{x}^{2}-\frac{5}{2}\text{x}-2$
AnswerTo add we proceed as follows:
$\big(\frac{7}{2}\text{x}^{3}-\frac{1}{2}\text{x}^{2}+\frac{5}{3}\big)+\big(\frac{3}{2}\text{x}^{3}+\frac{7}{4}\text{x}^{2}-\text{x}+\frac{1}{3}\big)\\+\big(\frac{3}{2}\text{x}^{2}-\frac{5}{2}\text{x}-2\big)$
$=\frac{7}{2}\text{x}^{3}-\frac{1}{2}\text{x}^{2}+\frac{5}{3}+\frac{3}{2}\text{x}^{3}+\frac{7}{4}\text{x}^{2}-\text{x}+\frac{1}{3}\\+\frac{3}{2}\text{x}^{2}-\frac{5}{2}\text{x}-2$
$=\frac{7}{2}\text{x}^{3}+\frac{3}{2}\text{x}^{3}-\frac{1}{2}+\frac{7}{4}\text{x}^{2}+\frac{3}{2}\text{x}^{2}-\text{x}$
$-\frac{5}{2}\text{x}+\frac{5}{3}\text{x}+\frac{5}{3}+\frac{1}{3}-2$ (Collecting like terms)
$=5\text{x}^{3}+\frac{11}{4}\text{x}^{2}-\frac{7}{2}\text{x}$ (combining like terms)
View full question & answer→Question 443 Marks
Find the values of the following expressions:
$81 x^2+16 y^2-72 x y$, when $x=\frac{2}{3}$ and $y=\frac{3}{4}$
AnswerLet us consider the following expression:
$81 x^2+16 y^2-72 x y$
Now
$81 x^2+16 y^2-72 x y=(9 x-4 y)^2$ (Using identity $(a+b)^2=a^2+2 a b+b^2$ )
$\Rightarrow 81 x^2+16 y^2-72 xy =\left[9\left(\frac{2}{3}\right)-4\left(\frac{3}{4}\right)\right]^2 \quad$ (Substituting $x=\frac{2}{3}$ and $y=\frac{3}{4}$ )
$\Rightarrow 81 x ^2+16 y ^2-72 xy =[6-3]^2$
$\Rightarrow 81 x ^2+16 y ^2-72 xy =3^2$
$\Rightarrow 81 x^2+16 y^2-72 x y=9$
View full question & answer→Question 453 Marks
Find the values of the following expressions:
$16 x^2+24 x+9$, when $x=\frac{7}{4}$
AnswerLet us consider the following expression:
$16 x^2+24 x+9$
Now
$16 x^2+24 x+9=(4 x+3)\left(\text { Using identity }(a+b)^2=a^2+2 a b+b^2\right)$
$16 x^2+24 x+9=\left(4 \times \frac{7}{4}+3\right)^2\left(\text { Substituting } x=\frac{7}{4}\right)$
$\Rightarrow 16 x^2+24 x+9=(7+3)^2$
$\Rightarrow 16 x^2+24 x+9=102$
$\Rightarrow 16 x^2+24 x+9=100$
View full question & answer→Question 463 Marks
Find following product:
$5 x^2 \times 4 x^3$
AnswerTo multiply algebraic expressions, we use commutative and associative laws along with the laws of indices. However, use of these laws are subject to their applicability in the given expressions.
In the present problem, to perform the multiplication, we can proceed as follows:
$5 x^2 \times 4 x^3$
$=(5 \times 4) \times\left(x^2 \times x^3\right)$
$=20 x^5\left(\because a^{m} \times a^{n}=a^{m+n}\right)$
Thus, the answer is $20 x^5$.
View full question & answer→Question 473 Marks
Find the product:
$x y\left(x^3-y^3\right)$
AnswerTo find the product, we will use distributive law as follows:
$x y\left(x^3-y^3\right)$
$=x y \times x^3-x y \times y^3$
$=\left(x \times x^3\right) \times y-x \times\left(y \times y^3\right)$
$=x^{1+3} y-x y^{1+3}$
$=x^4 y-x y^4$
Thus, the answer is $x^4 y-x y^4$
View full question & answer→Question 483 Marks
Simplify following:
[5 - 3x + 2y - (2x - y)] - (3x - 7y + 9)
Answer[5 - 3x + 2y - (2x- y)] - (3x - 7y + 9)
= [5 - 3x + 2y - 2x + y] - (3x - 7y + 9)
= [5 - 5x + 3y] - (3x - 7y + 9)
= 5 - 5x + 3y - 3x + 7y - 9
= 5 - 9 - 5x - 3x + 3y + 7y
= -4 - 8x +10y
View full question & answer→Question 493 Marks
Find the product:
$\frac{6\text{x}}{5}\big(\text{x}^3+\text{y}^3)$
AnswerTo find the product, we will use distributive law as follows:
$\frac{6\text{x}}{5}\big(\text{x}^3+\text{y}^3)$
$=\frac{6\text{x}}{5}×\text{x}^3+\frac{6\text{x}}{5}×\text{y}^3$
$=\frac{6}{5}×\big(\text{x}×\text{x}^3)+\frac{6}{5}×\big(\text{x}×\text{y}^3\big)$
$=\frac{6}{5}×\big(\text{x}\times\text{x}^{1+3}\big)+\frac{6}{5}×\big(\text{x}×\text{y}^3)$
$=\frac{6\text{x}^4}{5}+\frac{6\text{xy}^3}{5}$
Thus, the answer is $\frac{6\text{x}^4}{5}+\frac{6\text{xy}^3}{5}$
View full question & answer→Question 503 Marks
Find following product:
$\big(−\frac{24}{25}\text{x}^3\text{z}\big)×\big(\frac{−15}{16}\text{xz}^2\text{y)}$
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^m \times a^n=a^{m+n}$.We have:
$\big(−\frac{24}{25}\text{x}^3\text{z}\big)×\big(\frac{−15}{16}\text{xz}^2\text{y)}$
$=\big(-\frac{24}{25}×\frac{-15}{16}\big)×\big(\text{x}^3×\text{x})×\big(\text{z}\times\text{z}^2\big)\times\text{y}$
$=\big(-\frac{24}{25}×-\frac{15}{16}\big)×\big(\text{x}^{3+1}\big)×\big(\text{z}^{1+2}\big)\times\text{y}$
$=\frac{9}{10}\text{x}^4\text{y}\text{z}^3$
Thus, the answer is $=\frac{9}{10}\text{x}^4\text{y}\text{z}^3$
View full question & answer→Question 513 Marks
Subtract:
$\frac{3}{2}\text{x}-\frac{5}{4}\text{y}-\frac{7}{2}\text{z}$ from $\frac{2}{3}\text{x}+\frac{3}{2}\text{y}-\frac{4}{3}\text{z}$
Answer$\big(\frac{2}{3}\text{x}+\frac{3}{2}\text{y}-\frac{4}{3}\text{z}\big)-\big(\frac{3}{2}\text{x}-\frac{5}{4}\text{y}-\frac{7}{2}\text{z}\big)$
$=\frac{2}{3}\text{x}+\frac{3}{2}\text{y}-\frac{4}{3}\text{z}-\frac{3}{2}\text{x}+\frac{5}{4}\text{y}+\frac{7}{2}\text{z}$
$=\frac{2}{3}\text{x}-\frac{3}{2}\text{x}+\frac{3}{2}\text{y}+\frac{5}{4}\text{y}-\frac{4}{3}\text{z}+\frac{7}{2}\text{z}$ (Collecting like terms)
$=-\frac{5}{6}\text{x}+\frac{11}{4}\text{y}+\frac{13}{6}\text{z}$ (Combining like terms)
View full question & answer→Question 523 Marks
Find following product:
$\frac{1}{4}\text{xy}×\frac{2}{3}\text{x}^2\text{yz}^2$
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^m \times a^n=a^{m+n}$.
We have:
$\frac{1}{4}\text{xy}×\frac{2}{3}\text{x}^2\text{yz}^2$
$=\big(\frac{1}{4}×\frac{2}{3}\big)×\big(\text{x}×\text{x}^2)×(\text{y}×\text{y})×\text{z}^2$
$=\big(\frac{1}{4}×\frac{2}{3}\big)×\big(\text{x}^{1+2}\big)×\big(\text{y}^{1+1}\big)×\text{z}^2$
$=\frac{1}{6}\text{x}^3\text{y}^2\text{z}^2$
Thus, the answer is $=\frac{1}{6}\text{x}^3\text{y}^2\text{z}^2$
View full question & answer→Question 533 Marks
Find following product:
$(-5 x y) \times\left(-3 x^2 y z\right)$
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a ^{ m } \times$ $a^n=a^{m+n}$ wherever applicable.
We have:
$(-5 x y) \times\left(-3 x^2 y z\right)$
$=\{(-5) \times(-3)\} \times\left(x \times x^2\right) \times(y \times y) \times z$
$=15 \times\left(x^{1+2}\right) \times\left(y^{1+1}\right) \times z=15 x^3 y^2 z$
Thus, the answer is $15 x^3 y^2 z$.
View full question & answer→Question 543 Marks
Multiply:
$(2 x+8) \text { by }(x-3)$
AnswerTo multiply, we will use distributive law as follows:
$(2 x+8) \text { by }(x-3)$
$=2 x(x-3)+8(x-3)$
$=(2 x \times x-2 x \times 3)+(8 x-8 \times 3)$
$=\left(2 x^2-6 x\right)+(8 x-24)$
$=2 x^2-6 x+8 x-24$
$=2 x^2+2 x-24$
Thus, the answer is $2 x^2+2 x-24$.
View full question & answer→Question 553 Marks
Subtract:
$\text{x}^{2}\text{y}-\frac{4}{5}\text{x}\text{y}^{2}+\frac{4}{3}\text{xy}$ from $\frac{2}{3}\text{x}^{2}\text{y}+\frac{3}{2}\text{xy}^{2}-\frac{1}{3}\text{xy}$
Answer$\big(\frac{2}{3}\text{x}^{2}\text{y}+\frac{3}{2}\text{xy}^{2}-\frac{1}{3}\text{xy}\big)-\big(\text{x}^{2}\text{y}-\frac{4}{5}\text{x}\text{y}^{2}+\frac{4}{3}\text{xy}\big)$
$=\frac{2}{3}\text{x}^{2}\text{y}+\frac{3}{2}\text{xy}^{2}-\frac{1}{3}\text{xy}-\text{x}^{2}\text{y}-\frac{4}{5}\text{x}\text{y}^{2}+\frac{4}{3}\text{xy}$
$=\frac{2}{3}\text{x}^{2}-\text{x}^{2}\text{y}+\frac{3}{2}\text{x}\text{y}^2+\frac{4}{5}\text{x}\text{y}^2-\frac{1}{3}\text{xy}-\frac{4}{3}\text{xy}$ (Collecting like terms)
$=-\frac{1}{3}\text{x}^{2}\text{y}+\frac{23}{10}\text{x}\text{y}^{2}-\frac{5}{3}\text{xy}$ (Combining like terms)
View full question & answer→Question 563 Marks
Find following product:
$\big(−\frac{7}{5}\text{xy}^2\text{z}\big)×\big(\frac{13}{3}\text{x}^2\text{yz}^2\big)$
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a ^{ m \times}$ $a^n=a^{m+n} \cdot$
We have:
$\big(−\frac{7}{5}\text{xy}^2\text{z}\big)×\big(\frac{13}{3}\text{x}^2\text{yz}^2\big)$
$=\big(-\frac{7}{5}×\frac{13}{3}\big)×\big(\text{x}×\text{x}^2)\\×(\text{y}^2×\text{y})×\big(\text{z}\times\text{z}^2)$
$=\big(-\frac{7}{5}×\frac{13}{3}\big)×\big(\text{x}^{1+2}\big)×\big(\text{y}^{2+1}\big)\\×\big(\text{z}^{1+2}\big)$
$=-\frac{91}{15}\text{x}^3\text{y}^3\text{z}^3$
Thus, the answer is $=-\frac{91}{15}\text{x}^3\text{y}^3\text{z}^3.$
View full question & answer→Question 573 Marks
Find the following products:
$\left(3 x^2-4 x y\right)\left(3 x^2-3 x y\right)$
AnswerHere, we will use the identity $(x-a)(x-b)=x^2-(a+b) x+a b$.
$\left(3 x^2-4 x y\right)\left(3 x^2-3 x y\right)$
$=\left(3 x^2\right)^2-\left(4 x y+3 x y\left(3 x^2\right)+4 x y \times 3 x y\right.$
$=9 x^4-\left(12 x^3 y+9 x^3 y\right)+12 x^2 y^2$
$=9 x^4-21 x^3 y+12 x^2 y^2$
View full question & answer→Question 583 Marks
Evaluate following when x = 2, y = −1.
$(2\text{xy})×\big(\frac{\text{x}^2\text{y}}{4}\big)×\big(\text{x}^2\big)×\big(\text{y}^2\big)$
AnswerTo multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a^m \times a^n=a^{m+n}$.
$(2\text{xy})×\big(\frac{\text{x}^2\text{y}}{4}\big)×\big(\text{x}^2\big)×\big(\text{y}^2\big)$
$=\big(2×\frac{1}{4}\big)×\big(\text{x}×\text{x}^2×\text{x}^2\big)×\big(\text{y}×\text{y}×\text{y}^2)$
$=\big(2×\frac{1}{4}\big)×\big(\text{x}^{1+2+2}\big)×\big(\text{y}^{1+1+2}\big)$
$=\frac{1}{2}\text{x}^5\text{y}^4$
$\therefore(2\text{xy})×\big(\frac{\text{x}^2\text{y}}{4}\big)×\big(\text{x}^2\big)×\big(\text{y}^2\big)=\frac{1}{2}\text{x}^5\text{y}^4$
Substituting x = 2 and y = -1 in the result, we get:
$\frac{1}{2}\text{x}^5\text{y}^4$
$=\frac{1}{2}(2)^5(−1)^4$
$=\frac{1}{2}×32×1$
$= 16$
Thus, the answer is $16.$
View full question & answer→Question 593 Marks
Find the product:
$-5 a(7 a-2 b)$
AnswerTo find the product, we will use distributive law as follows:
$-5 a(7 a-2 b)$
$=(-5 a) \times 7 a+(-5 a) \times(-2 b)$
$=(-5 \times 7) \times(a \times a)+(-5 \times(-2)) \times(a \times b)$
$=(-35) \times\left(a^{1+1}\right)+(10) \times(a \times b)$
$=-35 a^2+10 a b$
Thus, the answer is $-35 a^2+10 a b$
View full question & answer→Question 603 Marks
$95 \times 105$
AnswerHere, we will use the identity $(a-b)(a+b)=a^2-b^2$
Let us consider the following expression:
$95 \times 105$
$\because \frac{95+105}{2}=\frac{200}{2}=100, \text { therefore, we will write the above product as: }$
$95 \times 105$
$=(100+5)(100-5)$
$=(100)^2-(5)^2$
$=10000-25$
$=9975$
Thus, the answer is 9975
View full question & answer→Question 613 Marks
Simplify:
$x^3 y\left(x^2-2 x\right)+2 x y\left(x^3-x^4\right)$
AnswerTo simplify, we will use distributive law as follows:
$x^3 y\left(x^2-2 x\right)+2 x y\left(x^3-x 4\right)$
$=x^5 y-2 x^4 y+2 x^4 y-2 x^5 y$
$=x^5 y-2 x^5 y-2 x^4 y+2 x^4 y$
$=-x^5 y$
View full question & answer→Question 623 Marks
If $x^2+y^2=29$ and $x y=2$, find the value of
$x^4+y^4$
AnswerWe have:
$\left(x^2+y^2\right)^2=x^4+2 x^2 y^2+y^4$
$\Rightarrow x^4+y^4=\left(x^2+y^2\right)^2-2 x^2 y^2$
$\Rightarrow x^4+y^4=\left(x^2+y 2\right)^2-2(x y)^2$
$\Rightarrow x^4+y^4=292-2(2)^2\left(\because x^2+y^2=29 \text { and } x y=2\right)$
$\Rightarrow x^4+y^4=841-8$
$\Rightarrow x^4+y^4=833$
View full question & answer→Question 633 Marks
Find the following products:
$(3 x-4 y)(2 x-4 y)$
AnswerHere, we will use the identity $(x-a)(x-b)=x^2-(a+b) x+a b$.
$(3 x-4 y)(2 x-4 y)$
$=(4 y-3 x)(4 y-2 x) \text { (Taking common - } 1 \text { from both parentheses) }$
$=(4 y)^2-(3 x+2 x)(4 y)+3 x \times 2 x$
$=16 y^2-(12 x y+8 x y)+6 x^2$
$=16 y^2-20 x y+6 x^2$
View full question & answer→Question 643 Marks
Add the following algebric expressions:
$4 x y^2-7 x^2 y, 12 x^2 y-6 x y^2,-3 x^2 y+5 x y^2$
AnswerTo add the like terms, we proceed as follows:
$\left(4 x y^2-7 x^2 y\right)+\left(12 x^2 y\right)-\left(6 x y^2\right)-\left(3 x^2 y+5 x y^2\right)$
$=4 x y^2-7 x^2 y+12 x^2 y-6 x y^2-3 x^2 y+5 x y^2$
$=4 x y^2-6 x y^2+5 x y^2-7 x^2 y+12 x^2 y-3 x^2 y \text { (Collecting like terms) }$
$=3 x^2 y+2 x^2 y \text { (Combining like terms) }$
View full question & answer→Question 653 Marks
Find the product:
$\big(−\frac{7}{4}\text{ab}^2\text{c}−\frac{6}{25}\text{a}^2\text{c}^2\big)\big(−50\text{a}^2\text{b}^2\text{c}^2\big)$
AnswerTo find the product, we will use distributive law as follows:$\big(−\frac{7}{4}\text{ab}^2\text{c}−\frac{6}{25}\text{a}^2\text{c}^2\big)\big(−50\text{a}^2\text{b}^2\text{c}^2\big)$
$=\Big\{\big(−\frac{7}{4}\text{ab}^2\text{c}\big)\big(−50\text{a}^2\text{b}^2\text{c}^2\big)\Big\}−\Big\{\big(\frac{6}{25}\text{a}^2\text{c}^2\big)(−50\text{a}^2\text{b}^2\text{c}^2\big)\Big\}$
$=\Big\{−\frac{7}{4}×(−50)\Big\}\big(\text{a}×\text{a}^2\big)×\big(\text{b}^2×\text{b}^2\big)×\big(\text{c}×\text{c}^2\big)\Big\}\\-\Big\{\big(\frac{6}{25}\big)(−50)\Big\}\big(\text{a}^2×\text{a}^2\big)×\big(\text{b}^2\big)×\big(\text{c}^2×\text{c}^2\big)\Big\}$
$=\Big\{−\frac{7}{4}×(−50)\Big\}\big(\text{a}^{1+2}\text{b}^{2+2}\text{c}^{1+2}\big)\\-\Big\{\big(\frac{6}{25}\big)(−50)\big(\text{a}^2+2\text{b}^2\text{c}^{2+2}\Big\}$
$=\frac{175}{2}\text{a}^3\text{b}^4\text{c}^3−\big(−12\text{a}^4\text{b}^2\text{c}^4\big)$
$=\frac{175}{2}\text{a}^3\text{b}^4\text{c}^3+12\text{a}^4\text{b}^2\text{c}^4$
Thus, the answer is $=\frac{175}{2}\text{a}^3\text{b}^4\text{c}^3+12\text{a}^4\text{b}^2\text{c}^4$
View full question & answer→Question 663 Marks
Find following product:
$(-7\text{xy})\times\big(\frac{1}{4}\text{x}^2\text{yz}\big)$
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^m \times$ $a^n=a^{m+n}$.
We have:
$(-7\text{xy})\times\big(\frac{1}{4}\text{x}^2\text{yz}\big)$
$=\big[\big(-7×\frac{1}{4}\big)\big]×\big(\text{x}×\text{x}^2\big)\times(\text{y}\times\text{y})\times\text{z}$
$=\big[\big(-7×\frac{1}{4}\big)\big]×\big(\text{x}^{\text{1+2}}\big)\times(\text{y}^{1+1}\big)\times\text{z}$
$=−\frac{7}{4}\text{x}^3\text{y}^2\text{z}.$
Thus, the answer is $−\frac{7}{4}\text{x}^3\text{y}^2\text{z}.$
View full question & answer→Question 673 Marks
Find the following product:
$2 a^3(3 a+5 b)$
AnswerTo find the product, we will use distributive law as follows:
$2 a^3(3 a+5 b)$
$=2 a^3 \times 3 a+2 a^3 \times 5 b$
$=(2 \times 3)\left(a^3 \times a\right)+(2 \times 5) a^3 b$
$=(2 \times 3) a^{3+1}+(2 \times 5) a^3 b$
$=6 a^4+10 a^3 b$
Thus, the answer is $6 a^4+10 a^3 b$.
View full question & answer→Question 683 Marks
If $x-y=7$ and $x y=9$, find the value of $x^2+y^2$
AnswerWe have:
$(x-y)^2=x^2-2 x y+y 2$
$\Rightarrow x^2+y^2=(x-y)^2+2 x y$
$\Rightarrow x^2+y^2=7^2+2 \times 9(\because x-y=7 \text { and } x y=9)$
$\Rightarrow x^2+y^2=49+18$
$\Rightarrow x^2+y^2=67$
View full question & answer→Question 693 Marks
What must be added to following expressions to make it a whole square?
$4 x^2-20 x+20$
AnswerLet us consider the following expression: $4 x^2-20 x+20$
The above expression can be written as:
$4 x^2-20 x+20=(2 x)^2-2 \times 2 x \times 5+20$
It is evident that if $2 x$ is considered as the first term and 5 is considered as the second term, 5 is required to be added to the above expression to make it a perfect square. Therefore, number 20 must become 25 .
Therefore, adding and subtracting 5 in the above expression, we get:
$\left(4 x^2-20 x+20+5\right)-5$
$=\left\{(2 x)^2-2 \times 2 x \times 5+20\right\}+5-5$
$=\left\{(2 x)^2-2 \times 2 x \times 5+25\right\}-5$
$=(2 x+5)^2-5$
Thus, the answer is 5 .
View full question & answer→Question 703 Marks
Find the product:
$-11 a(3 a+2 b)$
AnswerTo find the product, we will use distributive law as follows:
$-11 a(3 a+2 b)$
$=(-11 a) \times 3 a+(-11 a) \times 2 b$
$=(-11 \times 3) \times(a \times a)+(-11 \times 2) \times(a \times b)$
$=(-33) \times\left(a^{1+1}\right)+(-22) \times(a \times b)$
$=-33 a^2-22 a b$
Thus, the answer is $-33 a^2-22 a b$
View full question & answer→Question 713 Marks
Subtract:
$\frac{\text{ab}}{7}-\frac{35}{3}\text{bc}+\frac{6}{5}\text{ac}$ from $\frac{3}{5}\text{bc}-\frac{4}{5}\text{ac}$
Answer$\big(\frac{3}{5}\text{bc}-\frac{4}{5}\text{ac}\big)-\big(\frac{\text{ab}}{7}-\frac{35}{3}\text{bc}+\frac{6}{5}\text{ac}\big)$
$=\frac{3}{5}\text{bc}-\frac{4}{5}\text{ac}-\frac{\text{ab}}{7}+\frac{35}{3}\text{bc}-\frac{6}{5}\text{ac}\big)$
$=\frac{3}{5}\text{bc}+\frac{35}{3}\text{bc}-\frac{4}{5}\text{ac}-\frac{6}{5}\text{ac}-\frac{\text{ab}}{7}$
(Collecting like terms)
$=\frac{184}{15}\text{bc}-2\text{ac}-\frac{\text{ab}}{7}$ (Collecting like terms)
View full question & answer→Question 723 Marks
Simplify:
$x^2\left(x^2+1\right)-x^3(x+1)-x\left(x^3-x\right)$
AnswerTo simplify, we will use distributive law as follows:
$x^2\left(x^2+1\right)-x^3(x+1)-x\left(x^3-x\right)$
$=x^4+x^2-x^4-x^3-x^4+x^2$
$=x^4-x^4-x^4-x^3+x^2+x^2$
$=-x^4-x^3+2 x^2$
View full question & answer→Question 733 Marks
Simplify:
$x(x+4)+3 x\left(2 x^2-1\right)+4 x^2+4$
AnswerTo simplify, we will use distributive law as follows:
$x(x+4)+3 x\left(2 x^2-1\right)+4 x^2+4$
$=x^2+4 x+6 x^3-3 x+4 x^2+4$
$=x^2+4 x^2+4 x-3 x+6 x^3+4$
$=5 x^2+x+6 x^3+4$
View full question & answer→Question 743 Marks
Multiply:
$\left(3 x^2+y^2\right) \text { by }\left(2 x^2+3 y^2\right)$
AnswerTo multiply, we will use distributive law as follows:
$\left(3 x^2+y^2\right) \text { by }\left(2 x^2+3 y^2\right)$
$=3 x^2\left(2 x^2+3 y^2\right)+y^2\left(2 x^2+3 y^2\right)$
$=6 x^4+9 x^2 y^2+2 x^2 y^2+3 y^4$
$=6 x^4+11 x^2 y^2+3 y^4$
Thus, the answer is $6 x^4+11 x^2 y^2+3 y^4$.
View full question & answer→Question 753 Marks
Find the product:
$75\text{x}^2\text{y}\big(\frac{3}{5}\text{xy}^2+\frac{2}{5}\text{x}\big)$
AnswerTo find the product, we will use distributive law as follows:$75\text{x}^2\text{y}\big(\frac{3}{5}\text{xy}^2+\frac{2}{5}\text{x}\big)$
$=\frac{7}{5}\text{x}^2\text{y}×\frac{3}{5}\text{xy}^2+\frac{7}{5}\text{x}^2\text{y}×\frac{2}{5}\text{x}$
$=\frac{21}{25}\text{x}^{2+1}\text{y}^{1+2}+\frac{14}{25}\text{x}^{2+1}\text{y}$
$=\frac{21}{25}\text{x}^3\text{y}^3+\frac{14}{25}\text{x}^3\text{y}$
Thus, the answer is $\frac{21}{25}\text{x}^3\text{y}^3+\frac{14}{25}\text{x}^3\text{y}$
View full question & answer→Question 763 Marks
Simplify the following using the identities:
$\frac{198×198−102×102}{96}$
AnswerLet us consider the following expression:
$\frac{198×198−102×102}{96}=\frac{198^2-102^2}{96}$
Using the identity $(a+b)(a-b)=a^2-b^2$
we get:
$\frac{198×198−102×102}{96}=\frac{198^2-102^2}{96}\\=\frac{(198+102)(198−102)}{96}$
$\Rightarrow\frac{198×198−102×102}{96}=\frac{(198+102)(198−102)}{96}$
$\Rightarrow\frac{198×198−102×102}{96}=\frac{300×96}{96}$
$\Rightarrow\frac{198×198−102×102}{96}=300$
Thus, the answer is 300.
View full question & answer→Question 773 Marks
Subtract the sum of $2 x-x^2+5$ and $-4 x-3+7 x^2$ from 5 .
AnswerWe have to subtract the sum of $\left(2 x-x^2+5\right)$ and $(-4 x-3+7 x)$ from 5 .
$5-\left\{\left(2 x-x^2+5\right)+\left(-4 x-3+7 x^2\right)\right\}$
$=5-\left(2 x-4 x-x^2+7 x^2+5-3\right)$
$=5-2 x-4 x-x^2+7 x^2+5-3$
$=5-5+3-2 x+4 x+x^2-7 x^2 \text { (Collecting like terms) }$
$=3+2 x-6 x^2 \text { (Combining like terms) }$
Thus, the answer is $3+2 x-6 x^2$.
View full question & answer→Question 783 Marks
Find following product:
$(7 a b) \times\left(-5 a b^2 c\right) \times\left(6 a b c^2\right)$
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a ^{ m } \times$$a^n=a^{m+n}$
We have:
$(7 a b) \times\left(-5 a b^2 c\right) \times\left(6 a b c^2\right)$
$=\{7 \times(-5) \times 6\} \times(a \times a \times a) \times\left(b \times b^2 \times b\right) \times\left(c \times c^2\right)$
$=\{7 \times(-5) \times 6\} \times\left(a^{1+1+1}\right) \times\left(b^{1+2+1}\right) \times\left(c^{1+2}\right)$
$=-210 a^3 b^4 c^3$
Thus, the answer is $=-210 a^3 b^4 c^3$.
View full question & answer→Question 793 Marks
Simplify:
$\left(x^3-2 x^2+5 x-7\right)(2 x-3)$
AnswerTo simplify, we will proceed as follows: $\left(x^3-2 x^2+5 x-7(2 x-3)\right.$
$=2 x\left(x^3-2 x^2+5 x-7\right)-3\left(x^3-2 x^2+5 x-7\right)$
$=2 x^4-4 x^3+10 x^2-14 x-3 x^3+6 x^2-15 x+21$
$=2 x^4-4 x^3-3 x^3+10 x^2+6 x^2-14 x-15 x+21$
$=2 x^4-7 x^3+16 x^2-29 x+21$
Thus, the answer is $2 x^4-7 x^3+16 x^2-29 x+21$.
View full question & answer→Question 803 Marks
Take away:
$\frac{\text{y}^3}{3}+\frac{7\text{y}^2}{3}+\frac{1}{2}\text{y}+\frac{1}{2}$ from $\frac{1}{3}−\frac{5}{3}\text{y}^2$
AnswerThe difference is given by:$\big(\frac{1}{3}−\frac{5}{3}\text{y}^2\big)-\big(\frac{\text{y}^3}{3}+\frac{7\text{y}^2}{3}+\frac{1}{2}\text{y}+\frac{1}{2}\big)$
$=\frac{1}{3}−\frac{5}{3}\text{y}^2-\frac{\text{y}^3}{3}-\frac{7\text{y}^2}{3}-\frac{\text{y}}{2}-\frac{1}{2}$
$=\frac{1}{3}−\frac{1}{2}-\frac{\text{y}}{2}-\frac{5}{3}\text{y}^2-\frac{7\text{y}^2}{3}-\frac{\text{y}^3}{3}$
$=\big(\frac{2-3}{6}\big)-\frac{\text{y}}{2}+\big(\frac{-5-7}{3}\big)\text{y}^2-\frac{\text{y}^3}{3}$ (Collecting like terms)
$=-\frac{1}{6}-\frac{\text{y}}{2}-4\text{y}^2-\frac{\text{y}^3}{3}$ (Combining like terms)
View full question & answer→Question 813 Marks
Multiply:
$\left(x^6-y^6\right) \text { by }\left(x^2+y^2\right)$
AnswerTo multiply, we will use distributive law as follows:
$\left(x^6-y^6\right) \text { by }\left(x^2+y^2\right)$
$=x^6\left(x^2+y^2\right)-y^6\left(x^2+y 2\right)$
$=\left(x^8+x^6 y^2\right)-\left(y^6 x^2+y^8\right)$
$=x^8+x^6 y^2-y^6 x^2-y^8$
Thus, the answer is $x^8+x^6 y^2-y^6 x^2-y^8$.
View full question & answer→Question 823 Marks
Simplify:
$3 a^2+2(a+2)-3 a(2 a+1)$
AnswerTo simplify, we will use distributive law as follows:
$3 a^2+2(a+2)-3 a(2 a+1)$
$=3 a^2+2 a+4-6 a^2-3 a$
$=3 a^2-6 a^2+2 a-3 a+4$
$=-3 a^2-a+4$
View full question & answer→Question 833 Marks
Find following product:
$-3 a^2 \times 4 b^4$
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a ^{ m } \times$ $a^n=a^{m+n}$ wherever applicable.
We have:
$-3 a^2 \times 4 b^4$
$=(-3 \times 4) \times\left(a^2 \times b^4\right)$
$=-12 a^2 b^4$
Thus, the answer is $-12 a^2 b^4$
View full question & answer→Question 843 Marks
Multiply:
$[-3 d+(-7 f)] \text { by }(5 d+f)$
AnswerTo multiply, we will use distributive law as follows:
${[-3 d+(-7 f)](5 d+f)}$
$=(-3 d)(5 d+f)+(-7 f)(5 d+f)$
$=\left(-15 d^2-3 d f\right)+\left(-35 d f-7 f^2\right)$
$=-15 d^2-3 d f-35 d f-7 f^2$
$=-15 d^2-38 d f-7 f^2$
Thus, the answer is $-15 d^2-38 d f-7 f^2$.
View full question & answer→Question 853 Marks
Simplify following:
$\frac{11}{2}\text{x}^2\text{y}-\frac{9}{4}\text{xy}^2+\frac{1}{4}\text{xy}-\frac{1}{14}\text{y}^2\text{x}\\+\frac{1}{15}\text{yx}^2+\frac{1}{2}\text{xy}$
Answer$\frac{11}{2}\text{x}^2\text{y}-\frac{9}{4}\text{xy}^2+\frac{1}{4}\text{xy}-\frac{1}{14}\text{y}^2\text{x}\\+\frac{1}{15}\text{yx}^2+\frac{1}{2}\text{xy}$
$=\frac{11}{2}\text{x}^2\text{y}+\frac{1}{15}\text{yx}^2-\frac{9}{4}\text{xy}^2\\-\frac{1}{14}\text{y}^2\text{x}\frac{1}{4}\text{xy}+\frac{1}{2}\text{xy}$ (Collecting like terms)
$=\big(\frac{165+2}{30}\big)\text{x}^2\text{y}+\big(\frac{−63−2}{28})\text{xy}^2+\big(\frac{1+2}{4}\big)\text{xy}$
$=\frac{167}{30}\text{x}^2\text{y}−\frac{65}{28}\text{x}\text{y}^2+\frac{3}{4}\text{xy}$ (Combining like terms)
View full question & answer→Question 863 Marks
Find product:
$\Big(\frac{−2}{7}\text{a}^4\Big)×\Big(\frac{−3}{4}\text{a}^2\text{b}\Big)×\Big(\frac{−14}{5}\text{b}^2\Big)$
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^m \times a^n=a^{m+n}$.
We have:
$\Big(\frac{−2}{7}\text{a}^4\Big)×\Big(\frac{−3}{4}\text{a}^2\text{b}\Big)×\Big(\frac{−14}{5}\text{b}^2\Big)$
$=\Big[\big(\frac{−2}{7}\big)×\big(\frac{−3}{4}\big)×\big(\frac{−14}{5}\big)\Big]×(\text{a}^4×\text{a}^2)×(\text{b}×\text{b}^2)$
$=−\Big(\frac{2}{7}×\frac{3}{4}×\frac{14}{5}\Big)×\text{a}^{4+2}×\text{b}^{1+2}$
$=−\frac{3}{5}\text{a}^6\text{b}^3$
Thus, the answer is $=−\frac{3}{5}\text{a}^6\text{b}^3$
View full question & answer→Question 873 Marks
Find the values of the following expressions:
$64 x^2+81 y^2+144 x y$
AnswerLet us consider the following expression:
$64 x^2+81 y^2+144 x y$
Now
$\left.64 x^2+81 y^2+144 x y=(8 x+9 y)^2\left(\text { using identify }(a+b)^2=a^2+2 a b+b^2\right)\right)$
$⇒64\text{x}^2+81\text{y}^2+144\text{xy}=[8(11)+9(43)]^2 $ $\big(\text{Substituting} \ \text{x}=11 \text{and} \ \text{y}=\frac{4}{3}\big)$
$⇒64\text{x}^2+81\text{y}^2+144\text{xy}=[88+12]^2$
$⇒64\text{x}^2+81\text{y}^2+144\text{xy}=100^2$
$⇒64\text{x}^2+81\text{y}^2+144\text{xy}=10000$
View full question & answer→Question 883 Marks
Simplify:
$\left(x^2-2 y^2\right)(x+4 y) x^2 y^2$
AnswerTo simplify, we will proceed as follows:
$\left(x^2-2 y^2\right)(x+4 y) x^2 y^2$
$=\left[x^2(x+4 y)-2 y^2(x+4 y)\right] x^2 y^2$
$=\left(x^3+4 x^2 y-2 x y^2-8 y^3\right) x^2 y^2$
$=x^5 y^2+4 x^4 y^3-2 x^3 y^4-8 x^2 y^5$
Thus, the answer is $x^5 y^2+4 x^4 y^3-2 x^3 y^4-8 x^2 y^5$.
View full question & answer→Question 893 Marks
Find following product:
$\big(-\frac{1}{27}\text{a}^2\text{b}^2\big)\times\big(\frac{9}{2}\text{a}^3\text{b}^2\text{c}^2\big)$
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^{ m } \times a ^{ n }= a ^{ m + n }$.
We have:
$\big(-\frac{1}{27}\text{a}^2\text{b}^2\big)\times\big(\frac{9}{2}\text{a}^3\text{b}^2\text{c}^2\big)$
$=\big[\big(-\frac{1}{27}×\frac{9}{2}\big)\big]×\big(\text{a}^2×\text{a}^3\big)×\big(\text{b}^2×\text{b}^2\big)×\text{c}^2$
$=\big[\big(-\frac{1}{27}×\frac{9}{2}\big)\big]\times\big(\text{a}^{2+3}\big)\times\big(\text{b}^{2+2}\big)\times\text{c}^2$
Thus, the answer is $−\frac{1}{6}\text{a}^5\text{b}^4\text{c}^2.$
View full question & answer→Question 903 Marks
Simplify the following using the identities:
$\frac{58^2−42^2}{16}$
AnswerLet us consider the following expression:
Using the identity $(a+b)(a-b)=a^2-b^2$
We get:
$\frac{58^2−42^2}{16}$
$\frac{58^2−42^2}{16}=\frac{(58+42)(58−42)}{16}$
$⇒\frac{58^2−42^2}{16}=\frac{100×16}{16}$
$⇒\frac{58^2−42^2}{16}=100$
Thus, the answer is 100.
View full question & answer→Question 913 Marks
Add the following algebric expressions:
$\frac{3}{2}\text{a}-\frac{5}{4}\text{b}+\frac{2}{5}\text{c},$$\frac{2}{3}\text{a}-\frac{7}{2}\text{b}+\frac{7}{2}\text{c},$$\frac{5}{3}\text{a}+\frac{5}{2}\text{b}-\frac{5}{4}\text{c},$
AnswerTo add the like terms, we proceed as follows:
$\big(\frac{3}{2}\text{a}-\frac{5}{4}\text{b}+\frac{2}{5}\text{c}\big)+\big(\frac{2}{3}\text{a}-\frac{7}{2}\text{b}+\frac{7}{2}\text{c}\big)\\+\big(\frac{5}{3}\text{a}+\frac{5}{2}\text{b}-\frac{5}{4}\text{c}\big)$
$=\frac{3}{2}\text{a}-\frac{5}{4}\text{b}+\frac{2}{5}\text{c}+\frac{2}{3}\text{a}-\frac{7}{2}\text{b}+\frac{7}{2}\text{c}\\+\frac{5}{3}\text{a}+\frac{5}{2}\text{b}-\frac{5}{4}\text{c}$
$=\frac{3}{2}\text{a}+\frac{2}{3}\text{a}+\frac{5}{3}\text{a}-\frac{5}{4}\text{b}-\frac{7}{2}\text{b}+\frac{5}{2}\text{b}\\+\frac{2}{5}\text{c}+\frac{7}{2}\text{c}-\frac{5}{4}\text{c}$ (Collecting like terms)
$=\frac{23}{6}\text{a}-\frac{9}{4}\text{b}+\frac{53}{20}\text{a}$ (Combining like terms)
View full question & answer→Question 923 Marks
$9.8 \times 10.2$
AnswerHere, we will use the identity $(a-b)(a+b)=a^2-b^2$
Let us consider the following expression:
$9.8 \times 10.2$
$\because \frac{9.8+10.2}{2}=\frac{20}{2}=10, \text { t }$
$9.8 \times 10.2$
$=(10-0.2)(10+0.2)$
$=(10)^2-(0.2)^2$
$=100-0.04$
$=99.96$
$\because \frac{9.8+10.2}{2}=\frac{20}{2}=10 \text {, therefore, we will write the above product as: }$
Thus, the answer 99.96
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