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3 years ago

WILL MARK BRAINLIEST!!!

Mathematics

2 answers:

user100 [1]3 years ago

7 0

Answer: v[5/7] represents the volume of air inside a rubber ball with radius 5/7 units, such that

Step-by-step explanation:

Given: The volume of air inside a rubber ball with radius r can be found using the function.

Here volume is a dependent variable on radius (independent variable).

Therefore, the value of v[5/7] represents the volume of air inside a rubber ball with radius 5/7 units.

Read more on Brainly.com - brainly.com/question/3963009#readmore

mylen [45]3 years ago

5 0

Answer:

we know that

The volume of a sphere (rubber ball) is equal to

V = 4/3pi r^3

where r is the radius of the sphere (rubber ball)

The volume in function notation is equal to

V(r) = pi r^3

in this problem we have

V(5/7)

that means

The radius of the rubber ball is 5/7 ft and the volume of the rubber ball is V(5/7) ft^3

<u>therefore, the answer is the option B</u>

the volume of the rubber ball when the radius equals 5/7 feet

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Is anybody else here to help me ??

Akimi4 [234]

Answer:

$\cot(x)+\cot(\frac{\pi}{2}-x)$

$\cot(x)+\tan(x)$

$\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}$

$\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]$

$\csc(x)[\frac{1}{\cos(x)}]$

$\csc(x)[\sec(x)]$

$\csc(x)[\csc(\frac{\pi}{2}-x)]$

$\csc(x)\csc(\frac{\pi}{2}-x)$

Step-by-step explanation:

I'm going to use $x$ instead of $\theta$ because it is less characters for me to type.

I'm going to start with the left hand side and see if I can turn it into the right hand side.

$\cot(x)+\cot(\frac{\pi}{2}-x)$

I'm going to use a cofunction identity for the 2nd term.

This is the identity: $\tan(x)=\cot(\frac{\pi}{2}-x)$ I'm going to use there.

$\cot(x)+\tan(x)$

I'm going to rewrite this in terms of $\sin(x)$ and $\cos(x)$ because I prefer to work in those terms. My objective here is to some how write this sum as a product.

I'm going to first use these quotient identities: $\frac{\cos(x)}{\sin(x)}=\cot(x)$ and $\frac{\sin(x)}{\cos(x)}=\tan(x)$

So we have:

$\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}$

I'm going to factor out $\frac{1}{\sin(x)}$ because if I do that I will have the $\csc(x)$ factor I see on the right by the reciprocal identity:

$\csc(x)=\frac{1}{\sin(x)}$

$\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

Now I need to somehow show right right factor of this is equal to the right factor of the right hand side.

That is, I need to show $\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}$ is equal to $\csc(\frac{\pi}{2}-x)$ .

So since I want one term I'm going to write as a single fraction first:

$\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}$

Find a common denominator which is $\cos(x)$ :

$\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}$

$\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}$

$\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}$

By the Pythagorean Identity $\cos^2(x)+\sin^2(x)=1$ I can rewrite the top as 1:

$\frac{1}{\cos(x)}$

By the quotient identity $\sec(x)=\frac{1}{\cos(x)}$ , I can rewrite this as:

$\sec(x)$

By the cofunction identity $\sec(x)=\csc(x)=(\frac{\pi}{2}-x)$ , we have the second factor of the right hand side:

$\csc(\frac{\pi}{2}-x)$

Let's just do it all together without all the words now:

$\cot(x)+\cot(\frac{\pi}{2}-x)$

$\cot(x)+\tan(x)$

$\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}$

$\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]$

$\csc(x)[\frac{1}{\cos(x)}]$

$\csc(x)[\sec(x)]$

$\csc(x)[\csc(\frac{\pi}{2}-x)]$

$\csc(x)\csc(\frac{\pi}{2}-x)$

7 0

3 years ago