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3 years ago

What is 5x + + 25x + 16 +

Mathematics

1 answer:

bazaltina [42]3 years ago

3 0

2(15x + 8) I hope this helps you

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Please Help me!!! The question is: Which graph shows the solution to y > x – 8?

FinnZ [79.3K]

Answer:

the answer is B

Step-by-step explanation:

7 0

3 years ago

What is the x-coordinate of the solution to the following system of equations?

tatyana61 [14]

Answer

X-2y = 2

3x + y = 6

Ox=-2

Ox=0

Ox=2

Ox=3

= 0

Step-by-step explanation:

easy

8 0

3 years ago

CAN SOMEONE HELP???????????

guapka [62]

Answer:

g(-1) = -1

g(-0.25) = 0

g(2) = 2

Step-by-step explanation:

6 0

3 years ago

Ace gained $ 176 by selling 32 meters cloth for $1496. Find his loss per meters she sold it for $1276

nasty-shy [4]

Answer:xcvxvxcvxcvxcvc

Step-by-step explanation:

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5 0

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:

The Derivative of a Function

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