All categories

2 years ago

Work out the volume of this sphere.

Mathematics

1 answer:

navik [9.2K]2 years ago

3 0

Answer:

Volume of the sphere ( V ) = 1436.02 cm³

Step-by-step explanation:

Explanation:-

Given that the radius of the sphere r = 7 cm

Volume of the sphere

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

$V = \frac{4}{3} \pi (7)^{3}$

$V = \frac{4}{3} X3.14 X (7)^{3} = 1436.02$

V = 1436.02 cm³

Volume of the sphere

V = 1436.02 cm³

You might be interested in

What is the degree of a polynomial that has roots of -2, 4i and -4i ?

likoan [24]

Answer:

Step-by-step explanation:

three

6 0

3 years ago

HELP IM TIMED which expression represents the average of three test scores

Blizzard [7]

C is the correct answer

5 0

3 years ago

Read 2 more answers

Ples help me find slant assemtotes

FrozenT [24]

A polynomial asymptote is a function $p(x)$ such that

$\displaystyle\lim_{x\to\pm\infty}(f(x)-p(x))=0$

$(y+1)^2=4xy\implies y(x)=2x-1\pm2\sqrt{x^2-x}$

Since this equation defines a hyperbola, we expect the asymptotes to be lines of the form $p(x)=ax+b$ .

Ignore the negative root (we don't need it). If $y=2x-1+2\sqrt{x^2-x}$ , then we want to find constants $a,b$ such that

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0$

We have

$\sqrt{x^2-x}=\sqrt{x^2}\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=|x|\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=x\sqrt{1-\dfrac1x}$

since $x\to\infty$ forces us to have $x>0$ . And as $x\to\infty$ , the $\dfrac1x$ term is "negligible", so really $\sqrt{x^2-x}\approx x$ . We can then treat the limand like

$2x-1+2x-ax-b=(4-a)x-(b+1)$

which tells us that we would choose $a=4$ . You might be tempted to think $b=-1$ , but that won't be right, and that has to do with how we wrote off the "negligible" term. To find the actual value of $b$ , we have to solve for it in the following limit.

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-4x-b)=0$

$\displaystyle\lim_{x\to\infty}(\sqrt{x^2-x}-x)=\frac{b+1}2$

We write

$(\sqrt{x^2-x}-x)\cdot\dfrac{\sqrt{x^2-x}+x}{\sqrt{x^2-x}+x}=\dfrac{(x^2-x)-x^2}{\sqrt{x^2-x}+x}=-\dfrac x{x\sqrt{1-\frac1x}+x}=-\dfrac1{\sqrt{1-\frac1x}+1}$

Now as $x\to\infty$ , we see this expression approaching $-\dfrac12$ , so that

$-\dfrac12=\dfrac{b+1}2\implies b=-2$

So one asymptote of the hyperbola is the line $y=4x-2$ .

The other asymptote is obtained similarly by examining the limit as $x\to-\infty$ .

$\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0$

$\displaystyle\lim_{x\to-\infty}(2x-2x\sqrt{1-\frac1x}-ax-(b+1))=0$

Reduce the "negligible" term to get

$\displaystyle\lim_{x\to-\infty}(-ax-(b+1))=0$

Now we take $a=0$ , and again we're careful to not pick $b=-1$ .

$\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-b)=0$

$\displaystyle\lim_{x\to-\infty}(x+\sqrt{x^2-x})=\frac{b+1}2$

$(x+\sqrt{x^2-x})\cdot\dfrac{x-\sqrt{x^2-x}}{x-\sqrt{x^2-x}}=\dfrac{x^2-(x^2-x)}{x-\sqrt{x^2-x}}=\dfrac
 x{x-(-x)\sqrt{1-\frac1x}}=\dfrac1{1+\sqrt{1-\frac1x}}$

This time the limit is $\dfrac12$ , so

$\dfrac12=\dfrac{b+1}2\implies b=0$

which means the other asymptote is the line $y=0$ .

4 0

3 years ago

The infinite sequence 1, 6, 15, 28, 45, ... exhibits which pattern? multiply the term number by 2, subtract 1 from the result, m

Amanda [17]

The correct response is ;
multiply the term number by 2, subtract 1 from the result, multiply by 2 times the term number, and divide the result by 2.
if term number is x,
the number in the pattern for xth term is ;
multiply term number by 2 - 2x
subtract 1 - 2x - 1
multiply by 2 times the term number - (2x-1) * 2x
divide by 2 - ((2x-1) * 2x)/2
so first term x=1
((2*1-1) * 2*1)/2
2/2 = 1
second term x = 2
((2*2-1) * 2*2)/2
(3*4)/2 = 6
third term x = 3
((2*3-1) * 2*3)/2
5*6/2 = 15
fourth term x= 4
((2*4-1) * 2*4)/2
56/2 = 28
Therefore the above response shows the correct pattern

5 0

3 years ago

Read 2 more answers

What is find the total area

kenny6666 [7]

You have to find the area of a shape.

7 0

3 years ago