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3 years ago

Please help me out with this

Mathematics

1 answer:

andre [41]3 years ago

5 0

Answer:

$\huge\boxed{\sqrt[3]{c^4}=c^\frac{4}{3}}$

Step-by-step explanation:

$\sqrt[n]{a^m}=a^\frac{m}{n}\\\\\text{therefore}\\\\\sqrt[3]{c^4}=c^\frac{4}{3}$

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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

X=8 in standard form Please help ASAP

Bas_tet [7]

Answer:

Well X=8 in standard form is NaN or 8 * 10 to the power of 0

8 0

2 years ago

Select the correct product. (y + z)3 y3 + z3 y3 + 3y2z + 3yz2 + z3 3y + 3z y3z3

zhenek [66]

(y + z)³ ( i believe you were trying to do the expression to the power of 3)

(y + z)³ = (y + z)(y + z)(y + z)

Use the FOIL method:
(y + z)(y + z) = y² + yz + yz + z²
Simplify
y² + 2yz + z²

plug in the answer gotten back into the expression

(y² + 2yz + z²)(y + z)

FOIL again, distribute each number to the other.

y²(y) = y³
y(2yz) = 2y²z
y(z²) = yz²
y²(z) = y²z
2yz(z) = 2yz²
z(z²) = z³

2y²z + 2yz² + y³ + z³ + y²z + yz²

Simplify: add all like variables

2y²z + y²z = 3y²z
2yz² + yz² = 3yz²

3y²z + 3yz² + y³ + z³ is your answer, or y^3 + 3y^2z + 3yz^2 + z^3

hope this helps

7 0

3 years ago

A perfect square can never have the following digit in its ones place.

Allisa [31]

Answer:

trapezium

Step-by-step explanation:

ģddvvhgddvbhjt

nygfgjuygfgh

4 0

3 years ago

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