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2 years ago

Helpppppppppp me pleaseeeee

Mathematics

1 answer:

Vlad1618 [11]2 years ago

6 0

Answer:

A = 16 13unit^2 + 9unit^2 = 5 √10 = 15.8 = 16

Step-by-step explanation:

For Pythagoras we square route

= x^2 = 13^13 + (one of the closer squares for x elimination)

x= 169 + 81 169 + 100 169 + 49 169+64 169 + 121

x = 250 or 269 or 218 or 233 or 290

x= 5√10 or 15.8 or √269 269 or √218 2.8 or √233 or 2.33 ( or √290 = ( √48+2) or 2.9

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2 years ago

Jaheem has a goal running a total of 125 miles this month. Each day that he ran, he ran 7 miles. Which expression could Jaheem u

VashaNatasha [74]

Answer:

a. 125 – 7d

Step-by-step explanation:

Let d be the number of days.

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3 years ago

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2 years ago

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Len [333]

Step-by-step explanation:

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3 years ago

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

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

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

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3 years ago