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3 years ago

What is the answer to this question i am very confused on how to solve this equation

Mathematics

1 answer:

Ilia_Sergeevich [38]3 years ago

5 0

Answer:

3,925 feet ^2

Step-by-step explanation:

75 ft

80*

A = pi*r^2

pi * 75^2 = 17662.5

80/360 = 2/9

17662.5 * (2/9) = 3925

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The perimeter of a rectangular field is

Andre45 [30]

346 divided by 2 would be 153. Subtract 92, and that would lead you with 61.
Therefore, the ultimate answer is 61.

8 0

3 years ago

PLEASE ANSWER THIS AS SOON AS POSSIBLE

monitta

Answer:

-3

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6) Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

2 years ago

A hotel has 480 rooms. Of these rooms, 4/5 are non-smoking. How many rooms are ... a. nonsmoking? b. smoking?

Anestetic [448]

Answer: 384
There are 384 smoke free rooms.

7 0

3 years ago

A survey was taken of students in math classes to find out how many hours per day students spend on social media. The survey res

Yakvenalex [24]

The question would be:

Which is the best measure of center for first period and why?

Mean, because there are no outliers that affect the center

Median, because there is 1 outlier that affects the center

Interquartile range, because there is 1 outlier that affects the center

I think the answer would be the fourth choice "Standard deviation, because there are no outliers that affect the center"

I hope my answer has come to your help. God bless and have a nice day ahead!

8 0

3 years ago