Question
Using variables ? ??? ? write any two equations whose solution is (0,2)
✓
Answer
Get the step-by-step solution for this question inside the Vidyadip app.
Get the answer in the appNeed a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In the year 2010 in the village there were 4000 people who were literate. Every year the number of literate people increases by 400. How many people will be literate in the year 2020?
Show that the points A(1, 2), B(1, 6), C(1 + 2 √3, 4) are vertices of an equilateral triangle.
Check whether points (3, 3), (-4, -1) and (3, -5) are the vertices of an isosceles triangle.
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
A black king.
A black king.
Find the coordinates of the midpoint of segment $Q R$, if $Q(2.5,-4.3)$ and $R(-1.5,2.7)$
→A park os of the shape of a circle of diameter 7m. it is surrounded by a path of width of 0.7m find the expenditure of cementing the path, if its cost is rs. 110 sq m.$\Big[\text{Use }\pi=\frac{22}{7}\Big]$
→The area of a sector of a circle of radius 5cm is $5\pi \text{ cm}^2.$ Find the angle contained by the sector.
→The sum of first n terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another A.P. whose first term is -30 and common difference is 8. Find n.
The ratio of the areas of two triangles with equal height is $3: 2$. The base of the larger triangle is 18 cm . Find the corresponding base of the smaller triangle.
→The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. ($\pi=3.14$)
Activity :
$r=10$ cm, $\theta=90^{\circ}$, $\pi=3.14.$
$A(P-AXB) = \frac{\theta}{360} \times ⬜$
$= \frac{⬜}{360} \times 3.14 \times 10^{2}$
$= \frac{1}{4} \times ⬜$
$A(P-AXB) = ⬜$ sq.cm.
→Activity :
$r=10$ cm, $\theta=90^{\circ}$, $\pi=3.14.$
$A(P-AXB) = \frac{\theta}{360} \times ⬜$
$= \frac{⬜}{360} \times 3.14 \times 10^{2}$
$= \frac{1}{4} \times ⬜$
$A(P-AXB) = ⬜$ sq.cm.