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Lostsunrise [7]
2 years ago
15

Which expression is equivalent to 2(8n)^4

Mathematics
1 answer:
Radda [10]2 years ago
5 0

Answer:

2* 8n* 8n* 8n*8n

Step-by-step explanation:

Break it down 2(8n)^4 because 4 is the exponent you will multiply 8n four times so it will be 8n*8n*8n*8n so add the 2 to it and it will be 2*8n*8n*8n*8n. Hope this help :)  

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How to calculate the three averages for the given data?
vichka [17]

Answer:

Add the numbers together and divide by the number of numbers. (The sum of values divided by the number of values). Arrange the numbers in order, find the middle number. (The middle value when the values are ranked).

4 0
2 years ago
HELP please!!<br> “Find the volume of the sphere rounded to the nearest hundredth
zvonat [6]

Answer:

904.32 cm^3

Step-by-step explanation:

The formula for the volume of a sphere is V=4/3πr³. Since r is given, we can plug that in for r. I'm assuming that we are using 3.14 for pi, so when we plug in all the values in the equation we get V = 4/3*3.14*6³, which solves out to 904.32.

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matrenka [14]

Check the picture below.

4 0
2 years ago
–10p = –11p − 15<br> What is the answer
tensa zangetsu [6.8K]

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Step-by-step explanation:

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−10p+11p=−11p−15+11p

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4 0
3 years ago
Read 2 more answers
Need help with Calculus 1 inverse trig functions
lidiya [134]

Answer:

\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}

Step-by-step explanation:

<u>The Derivative of a Function</u>

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}

There are tables where the derivative of all known functions are provided for an easy calculation of specific functions.

The derivative of the inverse tangent is given as

\displaystyle (tan^{-1}u)'=\frac{u'}{1+u^2}

Where u is a function of x as provided:

y=3tan^{-1}(x+\sqrt{1+x^2})

If we set

u=(x+\sqrt{1+x^2})

Then

\displaystyle u'=1+\frac{2x}{2\sqrt{1+x^2}}

\displaystyle u'=1+\frac{x}{\sqrt{1+x^2}}

Taking the derivative of y

y'=3[tan^{-1}(x+\sqrt{1+x^2})]'

Using the change of variables

\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}

\displaystyle y'=3\frac{u'}{1+u^2}=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+(x+\sqrt{1+x^2})^2}

Operating

\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+x^2+2x\sqrt{1+x^2}+1+x^2}

\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}

8 0
3 years ago
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