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3 years ago

What are the three theorems we can use

Mathematics

1 answer:

Veronika [31]3 years ago

3 0

Answer:

Hope this will help you....

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Please help me out please please !!!!!

Sauron [17]

The surface area of a sphere is given by

$S = 4\pi r^2$

We deduce

$r = \sqrt{\dfrac{S}{4\pi}}$

So, in your case, the radius is

$r = \sqrt{\dfrac{100\pi}{4\pi}}=\sqrt{25}=5$

The volume of a sphere is given by

$V=\dfrac{4}{3}\pi r^3$

So, we have

$V = \dfrac{4}{3} \pi 5^3 = \dfrac{500}{3}\pi$

5 0

3 years ago

Find the value of a square + b square when a+b = 6 a_b = 4

Vanyuwa [196]

Answer:

a+b=6

a-b =4

______subtract

(a-a)+(b-(-b))=6-4

0+2b = 2

b= 1

put the value of b at any equation

a+1 =6

then a=6-1=5

a square+b square= 25+1= 26

4 0

3 years ago

Please help me thank you

Hoochie [10]

Answer:

$\large\boxed{\sin2\theta=\dfrac{\sqrt3}{2},\ \cos2\theta=\dfrac{1}{2}}$

Step-by-step explanation:

We know:

$\sin2\theta=2\sin\theta\cos\theta\\\\\cos2\theta=\cos^2\theta-\sin^2\thet$

We have

$\sin\theta=\dfrac{1}{2}$

Use $\sin^2\theta+\cos^2\theta=1$

$\left(\dfrac{1}{2}\right)^2+\cos^2\theta=1\\\\\dfrac{1}{4}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{1}{4}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{4}{4}-\dfrac{1}{4}\\\\\cos^2\theta=\dfrac{3}{4}\to\cos\theta=\pm\sqrt{\dfrac{3}{4}}\to\cos\theta=\pm\dfrac{\sqrt3}{\sqrt4}\to\cos\theta=\pm\dfrac{\sqrt3}{2}\\\\\theta\in[0^o,\ 90^o],\ \text{therefore all functions have positive values or equal 0.}\\\\\cos\theta=\dfrac{\sqrt3}{2}$

$\sin2\theta=2\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt3}{2}\right)=\dfrac{\sqrt3}{2}\\\\\cos2\theta=\left(\dfrac{\sqrt3}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{3-1}{4}=\dfrac{2}{4}=\dfrac{1}{2}$