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katen-ka-za [31]
3 years ago
12

An expression is equivalent to

Mathematics
1 answer:
Phoenix [80]3 years ago
3 0
I think that is for your need

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I need help with this
m_a_m_a [10]
Nearest 100: 543,800
Nearest 1,000: 544,800
Nearest 10,000: 540,800

6 0
3 years ago
Read 2 more answers
Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20
Mice21 [21]

Power and chain rule (where the power rule kicks in because \sqrt x=x^{1/2}):

\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'

Simplify the leading term as

\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}

Quotient rule:

\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}

Chain rule:

(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)

(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)

Put everything together and simplify:

\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}

=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}

=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}

5 0
3 years ago
How do i total this please help<br> 8.3m<br> 5.2m<br> 4.36m<br> 42.1m<br> 86.1m
I am Lyosha [343]
Add them all up so it would equal 146.06
3 0
2 years ago
Explain the process for finding the product of two integers
velikii [3]
When you multiply two integers, or in fact any two numbers, you are finding the product of those two numbers;
So finding the product is done through multiplication.
5 0
3 years ago
Write parametric equations of the line y=4x-5
Rasek [7]

Answer:

A set of parametric equations for the line y = 4x - 5 is;

x = t

y = 4t - 5

Step-by-step explanation:

To find a set of parametric equations for the line y = 4x - 5;

We can assign either variable x or y equal to the parameter t, in this case we can easily let x = t

We then substitute x = t in the original equation;

y = 4t - 5

Therefore, a set of parametric equations for the line y = 4x - 5 is;

x = t

y = 4t - 5

7 0
3 years ago
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