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2 years ago

Simplify the expression: 2d=1x+5b+1x=6p

Mathematics

2 answers:

Naya [18.7K]2 years ago

8 0

Answer:

2d+1x+5b+1x=6p

Step-by-step explanation:

Rewrite = so is on the left side:

2d+1x+5b+1x=6p

Hence, the correct answer is 2d+1x+5b+1x=6p

Ilya [14]2 years ago

5 0

Answer:

2d = 2x + 5b = 6p

Step-by-step explanation:

2d = 1x + 5b + 1x = 6p

2d = 2x + 5b = 6p

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_____ + 5 3/4 = 9 5/8<br>please help you will get 75 points

LuckyWell [14K]

Answer:

3 7/8

Step-by-step explanation:

_____ + 5 3/4 = 9 5/8

x + 5 3/4 = 9 5/8

Subtract 5 3/4 from each side

x + 5 3/4 - 5 3/4 = 9 5/8 - 5 3/4

x = 9 5/8 - 5 3/4

Get a common denominator of 8

5 3/4 *2/2 = 5 6/8

x = 9 5/8 - 5 6/8

We need to borrow 1 from the 9 in the form of 8/8

x= 8 + 8/8+ 5/8 - 5 6/8

x = 8 13/8 - 5 6/8

= 3 7/8

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You and your family took a trip over track out that was 1.5 hours away. Which two trips could your family have taken?

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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