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Roots are distinct and real when b2 - 4ac = 5, not real when b2 - 4ac = -5.
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To check the rule for the terms of the sequence look at the arrangements and fill the empty boxes suitably.
1,4,7,10,13,…
1,4,7,10,13,…
Smita has invested ₹ 12,000 and purchased shares of FV ₹ 10 at a premium of ₹ 2. Find the number of shares she purchased. complete the given activity to get the answer.
FV = ₹ 10, Premium = ₹ 2
∴ MV = FV + ⬜
= ⬜ + ⬜ = ⬜
$\therefore \quad$ No. of shares $=\frac{\text { Totalinvestment }}{ MV }=$ = $=\frac{12000}{⬜}$
= ⬜ shares
Smita has purchased ⬜ shares.
→FV = ₹ 10, Premium = ₹ 2
∴ MV = FV + ⬜
= ⬜ + ⬜ = ⬜
$\therefore \quad$ No. of shares $=\frac{\text { Totalinvestment }}{ MV }=$ = $=\frac{12000}{⬜}$
= ⬜ shares
Smita has purchased ⬜ shares.
First term and common difference of an A.P. are 6 and 3 respectively ; find S27.
$a=6, d=3, S_{27}=?$
$S_n=\frac{n}{2}$[⬜ + (n - 1)d]
$\left.\therefore \quad S _{27}=\frac{27}{2}[12+(27-1) ⬜\right]$
$=\frac{27}{2} \times ⬜$
$=27 \times 45$ = ⬜
→$a=6, d=3, S_{27}=?$
$S_n=\frac{n}{2}$[⬜ + (n - 1)d]
$\left.\therefore \quad S _{27}=\frac{27}{2}[12+(27-1) ⬜\right]$
$=\frac{27}{2} \times ⬜$
$=27 \times 45$ = ⬜
Shri Shantilal purchased 150 shares of FV ₹ 100, for MV ₹ 120. Company paid dividend 7% Complete the following activity to find the rate of return on his investment.
FV =\%100; Number of shares =? 150 ; MV = mathbb * 120
FV =\%100; Number of shares =? 150 ; MV = mathbb * 120
Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm. Complete the following activity.
Activity: As shown in figure GLMNT is a reactangle.
$\therefore$ Area of rectangle $=$ length $\times$ breadth
$\therefore$ Area of rectangle $=\square \times$ breadth
$\therefore 192=\square \times$ breadth
$\therefore$ Breadth $=12 cm$
Also,
$\angle TLM =90^{\circ}$ [Each angle of reactangle is right angle]
In $\triangle T L M$,
By Pythagoras theorem
$\therefore TM ^2= TL ^2+\square$
$\therefore TM ^2=12^2+\square$
$\therefore TM ^2=144+\square$
$\therefore TM ^2=400$
$\therefore TM =20$
→Activity: As shown in figure GLMNT is a reactangle.
$\therefore$ Area of rectangle $=$ length $\times$ breadth
$\therefore$ Area of rectangle $=\square \times$ breadth
$\therefore 192=\square \times$ breadth
$\therefore$ Breadth $=12 cm$
Also,
$\angle TLM =90^{\circ}$ [Each angle of reactangle is right angle]
In $\triangle T L M$,
By Pythagoras theorem
$\therefore TM ^2= TL ^2+\square$
$\therefore TM ^2=12^2+\square$
$\therefore TM ^2=144+\square$
$\therefore TM ^2=400$
$\therefore TM =20$
Complete the following activity to form a quadratic equation.
Activity :
Activity :
The first term of an A.P. is 5 and the common difference is 4 . Complete the following activity to find the sum of first 12 terms of this A.P.
Write the correct number in the given boxes from the following A. P.
70, 60, 50, 40, . . .
Here
$t_1$ = ⬜, $t_2$ = ⬜, $t_3$ = ⬜, ....
∴ a = ⬜, d = ⬜
→70, 60, 50, 40, . . .
Here
$t_1$ = ⬜, $t_2$ = ⬜, $t_3$ = ⬜, ....
∴ a = ⬜, d = ⬜
Complete the following activity to find the length of hypotenuse of right angled triangle, if sides of right angle are $9 cm$ and $12 cm$.
Activity: In $\triangle P Q R, m \angle P Q R=90^{\circ}$
By Pythagoras Theorem,
$P Q^2+\square=P R^2$
$\therefore P R^2=9^2+12^2$
$\therefore P R^2=\square+144$
$\therefore P R^2=\square$
$\therefore P R=15$
$\therefore$ Length of hypotenuse of triangle PQR is $\square cm$.
→Activity: In $\triangle P Q R, m \angle P Q R=90^{\circ}$
By Pythagoras Theorem,
$P Q^2+\square=P R^2$
$\therefore P R^2=9^2+12^2$
$\therefore P R^2=\square+144$
$\therefore P R^2=\square$
$\therefore P R=15$
$\therefore$ Length of hypotenuse of triangle PQR is $\square cm$.
Write the values of the following trigonometric ratios.
$\tan 30^{\circ}=\frac{⬜}{⬜}$
→$\tan 30^{\circ}=\frac{⬜}{⬜}$