Question
To solve simultaneous equations $3 x+2 y=29$ and $10 x-2 y=36$.
Adding the given equations,
Adding the given equations,
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Answer
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Similar questions
In fig. BP ⊥ AC, CQ ⊥ AB, A−P−C, and A−Q−B then show that ΔAPB and ΔAQC are similar.
In ΔAPB and ΔAQC
In ΔAPB and ΔAQC
∠APB = [ ]° ......(i)
∠AQC = [ ]° ......(ii)
∠APB ≅ ∠AQC .....[From (i) and (ii)]
∠PAB ≅ ∠QAC .....[______]
ΔAPB ~ ΔAQC .....[______]
Complete the following table using given information.
Form a ‘Road safety commitee’ of two, from 2 boys (B1, B2) and 2 girls(G1, G2). Complete the following activity to write the sample space.
(a) Committee of 2 boys = ⬜
(b) Committee of 2 girls = ⬜
(c) Committee of one boy and one girl = $B_1 G_1$, ⬜, ⬜, ⬜
∴ Sample space = {..., ..., ..., ..., ..., ...}
→(a) Committee of 2 boys = ⬜
(b) Committee of 2 girls = ⬜
(c) Committee of one boy and one girl = $B_1 G_1$, ⬜, ⬜, ⬜
∴ Sample space = {..., ..., ..., ..., ..., ...}
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 correct number in the given boxes from the following A. P.
– 3, – 8, – 13, – 18, . . .
$\text { Here } t _3=\square, t _2=\square, t _4=\square, t _1=\square $
$ t _2- t _1=\square, t _3- t _2=\square $
$ \therefore a =\square, d =\square$
→– 3, – 8, – 13, – 18, . . .
$\text { Here } t _3=\square, t _2=\square, t _4=\square, t _1=\square $
$ t _2- t _1=\square, t _3- t _2=\square $
$ \therefore a =\square, d =\square$
In the adjoining figure, AB || CD || EF. If AC = 5.4, CE = 9, BD = 7.5, then find DF.
$AB \| CD \| EF$
$\therefore \frac{ AC }{ ⬜ }=\frac{ ⬜ }{ DF }$
$\therefore \quad \frac{5.4}{9}=\frac{⬜}{ DF }$
$\therefore \quad DF =\frac{7.5 \times 9}{5.4}$
∴ DF = ⬜
[Given]
⬜
→$AB \| CD \| EF$
$\therefore \frac{ AC }{ ⬜ }=\frac{ ⬜ }{ DF }$
$\therefore \quad \frac{5.4}{9}=\frac{⬜}{ DF }$
$\therefore \quad DF =\frac{7.5 \times 9}{5.4}$
∴ DF = ⬜
[Given]
⬜
In $\triangle A B C$, ray $B D$ bisects $\angle A B C . A-D-C$, side $D E \|$ side $B C, A-E-B$, then prove that $\frac{A B}{B C}=\frac{A E}{E B}$
In $\triangle A B C$, ray $B D$ bisects $\angle B$.
$\therefore \quad \frac{ AB }{ BC }=\frac{ AD }{ DC }$
In $\triangle ABC , DE \| BC$
$\therefore \quad \frac{ AE }{ EB }=\frac{ AD }{ DC }$
$\therefore \frac{ AB }{ ⬜ }=\frac{ ⬜ }{ EB }$
[Given]
[Angle biscetor theorem]
[Given]
⬜
[From (i) and (ii)]
→In $\triangle A B C$, ray $B D$ bisects $\angle B$.
$\therefore \quad \frac{ AB }{ BC }=\frac{ AD }{ DC }$
In $\triangle ABC , DE \| BC$
$\therefore \quad \frac{ AE }{ EB }=\frac{ AD }{ DC }$
$\therefore \frac{ AB }{ ⬜ }=\frac{ ⬜ }{ EB }$
[Given]
[Angle biscetor theorem]
[Given]
⬜
[From (i) and (ii)]
There are 9 tickets in a box, each bearing one of the numbers from 1 to 9 . One ticket is drawn at random from the box.
Event A : Ticket shows an even number.
Complete the following activity :
Event A : Ticket shows an even number.
Complete the following activity :
Fill in the blanks with correct number
$\left|\begin{array}{ll}
3 & 2 \\
4 & 5
\end{array}\right|=3 \times \square-\square \times 4=\square-8=\square$
→$\left|\begin{array}{ll}
3 & 2 \\
4 & 5
\end{array}\right|=3 \times \square-\square \times 4=\square-8=\square$
Complete the following activity to find the sum of natural numbers from 1 to 140 which are divisible by 4.
Sum of numbers from 1 to 140, which are divisible by 4 = $\square$
→Sum of numbers from 1 to 140, which are divisible by 4 = $\square$