Question
Factorise : $80-5 a^2$
✓
Answer
$5(4+a)(4-a)$
Need 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
Find four rational numbers between: 4 and 4.5.
[Hint. Here $a=4, b=4.5$ and $n=4$. Use $d=\frac{(b-a)}{(n+1)}=\frac{(4.5-4)}{(4+1)}=\frac{0.5}{5}=0.1$
So, the required numbers are a + d, a + 2d, a + 3d, a + 4d, i.e., 4.1, 4.2, 4.3 and 4.4]
→[Hint. Here $a=4, b=4.5$ and $n=4$. Use $d=\frac{(b-a)}{(n+1)}=\frac{(4.5-4)}{(4+1)}=\frac{0.5}{5}=0.1$
So, the required numbers are a + d, a + 2d, a + 3d, a + 4d, i.e., 4.1, 4.2, 4.3 and 4.4]
Express in terms of $\log 2$ and $\log 3 :\log \frac{26}{51}-\log \frac{91}{119}$
→$\triangle PQR$ is isosceles with $PQ = PR. T$ is the mid$-$point of $QR,$ and $TM$ and $TN$ are perpendiculars on $PR$ and $PQ$ respectively. Prove that,$PT$ is the bisector of $\angle P.$
→ The width of a rectangular room is $\frac{4}{7}$ of its length, $x,$ and its perimeter is $y.$ Write an equation connecting $x$ and $y.$ Find the length of the room when the perimeter is $4400 \ cm.$
→Solve for $x :\frac{\log 225}{\log 15}=\log x$
→The cross-section of a tunnel, perpendicular to its length is a trapezium ABCD in which AB = 8m DC 6 m and AL = BM The height of the tunnel is 2.4 m and its length is 40 m.
Find:
(i) the cost of paving the floor of the tunnel at 16 per m ^ 2
(ii) the cost of painting the internal surface of the tunnel, excluding the floor at the rate of 5 per $m ^2$
→Find:
(i) the cost of paving the floor of the tunnel at 16 per m ^ 2
(ii) the cost of painting the internal surface of the tunnel, excluding the floor at the rate of 5 per $m ^2$
The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. Also, find the cost of cultivating the field at ₹ 24.60 per 100$m ^2$.
→
If $\cos \theta=\frac{2 x}{1+x^2}$, find the values of sin $\theta$ and tan $\theta$ in terms of x.
$\left[\right.$ Hint. $\cos \theta=\frac{\text { Base }}{H y p .}=\frac{A B}{A C}=\frac{2 x}{\left(1+x^2\right)}$
Perp. $B C=\sqrt{A C^2-A B^2}=\sqrt{\left(1+x^2\right)^2-4 x^2}=\sqrt{x^4-2 x^2+1}=\left(x^2-1\right)$
$\left.\therefore \sin \theta=\frac{\text { Perp. }}{\text { Hyp. }}=\frac{\left(x^2-1\right)}{\left(1+x^2\right)} ; \tan \theta=\frac{\text { Perp. }}{\text { Base }}=\frac{\left(x^2-1\right)}{2 x}\right]$
Draw the graph for the linear equation given below$:2x - 3y = 4$
→→