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, $a^m× a^n= a^{m+n}$ 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 $15x^3y^2z.$
View full question & answer→Question 543 Marks
Multiply:
$(2x + 8) by (x - 3)$
AnswerTo multiply, we will use distributive law as follows: $(2x + 8) by (x - 3)$
$ = 2x(x - 3) + 8(x - 3) $
$= (2x × x - 2x × 3) + (8x - 8 × 3) $
$= (2x^2- 6x) + (8x - 24) $
$= 2x^2- 6x + 8x - 24$
$= 2x^2+ 2x - 24$
Thus, the answer is $2x^2+ 2x - 24.$
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, $a^m \times a^n=a^{m+n}$.
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:
$(3x^2- 4xy)(3x^2- 3xy)$
AnswerHere, we will use the identity $(x - a)(x - b) = x^2- (a + b)x + ab.$
$ \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.$a^m \times a^n=a^{m+n}$.
We have:
$(2\text{xy})×\big(\frac{\text{x}^2\text{y}}{4}\big)×\big(\text{x}^2\big)×\big(\text{y}^2\big)$
$=\big(2×\frac{1}{4}\big)×\big(\text{x}×\text{x}^2×\text{x}^2\big)×\big(\text{y}×\text{y}×\text{y}^2)$
$=\big(2×\frac{1}{4}\big)×\big(\text{x}^{1+2+2}\big)×\big(\text{y}^{1+1+2}\big)$
$=\frac{1}{2}\text{x}^5\text{y}^4$
$\therefore(2\text{xy})×\big(\frac{\text{x}^2\text{y}}{4}\big)×\big(\text{x}^2\big)×\big(\text{y}^2\big)=\frac{1}{2}\text{x}^5\text{y}^4$
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:
$-5a(7a - 2b)$
AnswerTo find the product, we will use distributive law as follows:
$-5a(7a - 2b) $
$= (-5a) × 7a + (-5a) × (-2b)$
$ = (-5 × 7) × (a × a) + (-5 × (-2)) × (a × b) $
$= (-35) × (a^{1+1}) + (10) × (a × b) $
$= -35a^2+ 10ab $
Thus, the answer is $-35a^2+ 10ab$
View full question & answer→Question 603 Marks
$95 × 105$
AnswerHere, we will use the identity $(a - b) (a + b) = a^2- b^2$
Let us consider the following expression:
$95 × 105$
$\because\frac{95+105}{2}=\frac{200}{2}=100,$
therefore, we will write the above product as:
$95 × 105 = (100 + 5) (100 - 5)$
$=(100)^2- (5)^2$
$= 10000 - 25$
$= 9975$
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 $xy = 2,$ 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:
$(3x - 4y)(2x - 4y)$
AnswerHere, we will use the identity $(x - a)(x - b) = x^2- (a + b)x + ab.$
$(3x - 4y)(2x - 4y)$
$= (4y - 3x)(4y - 2x)$ (Taking common $- 1$ from both parentheses)
$= (4y)^2- (3x + 2x)(4y) + 3x × 2x$
$= 16y^2- (12xy + 8xy) + 6x^2$
$= 16y^2- 20xy + 6x^2$
View full question & answer→Question 643 Marks
Add the following algebric expressions:
$ \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) $
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 $
$=4 x y^2-6 x y^2+5 x y^2-7 x^2 y+12 x^2 y-3 x^2 y$ (Collecting like terms)
$=3 x^2 y+2 x^2 y$ (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 $xy = 9$, 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?
$4x^2- 20x + 20$
AnswerLet us consider the following expression:$4x^2- 20x + 20$
The above expression can be written as:
$4x^2- 20x + 20 = (2x)^2- 2 × 2x × 5 + 20$
It is evident that if $2x$ 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: $\frac{198\times 198−102\times 102}{96}$
AnswerLet us consider the following expression:
$\frac{198\times 198−102\times 102}{96}=\frac{198^2-102^2}{96}$ Using the identity $(a+b)(a-b)=a^2-b^2$
we get:
$\frac{198\times 198−102\times 102}{96}=\frac{198^2-102^2}{96}$
$=\frac{(198+102)(198−102)}{96}$
$\Rightarrow\frac{198\times 198−102\times 102}{96}=\frac{(198+102)(198−102)}{96}$
$\Rightarrow\frac{198\times 198−102\times 102}{96}=\frac{300\times 96}{96}$
$\Rightarrow\frac{198\times 198−102\times 102}{96}=300$
Thus, the answer is $300.$
View full question & answer→Question 773 Marks
Subtract the sum of $(2x - x^2+ 5)$ and $(-4x - 3 + 7x^2)$ from $5.$
AnswerWe have to subtract the sum of $(2x - x^2+ 5)$ and $(-4x - 3 + 7x)$ 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, $a^m× a^n = a^{m+n}$.
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(2 x-3)\right. $
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)](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:
$64 x^2+81 y^2+144 x y=(8 x+9 y)^2 ($using identify $\left.(a+b)^2=a^2+2 a b+b^2\right) )$
$⇒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
$9.8 × 10.2$
AnswerHere, we will use the identity $(a - b) (a + b) = a^2- b^2$
Let us consider the following expression:
$9.8 × 10.2$
$\because\frac{9.8+10.2}{2}=\frac{20}{2}=10,$therefore, we will write the above product as:
$9.8 × 10.2$
$= (10 - 0.2) (10 + 0.2)$
$= (10)^2- (0.2)^2$
$= 100 - 0.04$
$= 99.96$
Thus, the answer $99.96$
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