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3 years ago

SOLVE 3 AND 4 I WILL GIVE BRAINLIST!!

Mathematics

1 answer:

ZanzabumX [31]3 years ago

7 0

Answer:

3. C≈31.42

4.C≈94.25

Step-by-step explanation:

10 ÷ 2 =5

C=2πr

c=2 (3.14) 5

c= 2(3.14)15

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Leila pumped 6 gallons of water out of her pool each minute for 24 minutes what was the total change in the amount of water in h

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Answer:

The answer is 144.

Step-by-step explanation:

The question said 6 gallons of water comes out every minute. So, if you multiply 6x24 you get the answer of 144.

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3 years ago

Which model shows 1/2 times 1/3

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Answer:

Where's the model?

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2 years ago

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Ab + c = d solve for B

Zepler [3.9K]

Answer:

b = (d-c)/a

Step-by-step explanation:

ab + c = d

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3 years ago

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77+ 2(3x – 5) = 8 – 3(2x + 1)<br> 78 2 + 3(2 -7) – 0 - 13 I 1).

nika2105 [10]

Answer:

-31/6

Step-by-step explanation:

77+2(3x-5)=8-3(2x+1)

77+6x-10=8-6x-3

77-10+6x=8-3-6x

67+6x=5-6x

67+6x-(-6x)=5

67+6x+6x=5

67+12x=5

12x=5-67

12x=-62

x=-62/12

simplify

x=-31/6

3 0

3 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> 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