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2 years ago

6,600,000,000 exponential notation

Mathematics

1 answer:

jeyben [28]2 years ago

8 0

Answer:

6.6 x 10^9

Step-by-step explanation:

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The first row of seats in Section G of a baseball stadium has 5 seats . There is a total of 20 rows of seats in the section . If

RoseWind [281]

Answer:

140

Step-by-step explanation:

8 0

2 years ago

−8 3/4÷2 1/6 please show me ur work

nalin [4]

Answer:

- 4 1/26

Step-by-step explanation:

<u>Solving in steps:</u>

- 8 3/4 ÷ 2 1/6 =
- 35/4 ÷ 13/6 =
- 35/4 × 6/13 =
- 35/2 × 3/13 =
- 105/26 =
- 4 1/26

6 0

3 years ago

The price ( p) of a school lunch is 8:50 The restaurant increases the price by 12% which statement is true for this situation

professor190 [17]

Answer:

Step-by-step explanation:

1 is 100% so increasing it by 12% means adding 0.12 to it

4 0

2 years ago

WILL GIVE BRAINLIEST

Ray Of Light [21]

Answer:

(x,y) --> (x, y-5)

Step-by-step explanation:

the y-intercept of f(x) = (0,2)

the y-intercept of g(x) = (0, -3)

-3 - 2 = -5

4 0

2 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

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

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

8 0

2 years ago