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2 years ago

Please help me with math

Mathematics

2 answers:

RSB [31]2 years ago

6 0

Answer:

These are the answers. Hope this helps!

Step-by-step explanation:

Mrac [35]2 years ago

4 0

IDisk dijo en la casa que se lo pase bien a mi madre que me pase la prueba y yo no me estaba por el verde y ni me siguió porque yo le voy a decir que no me puedes decir si me estaba llamando para allá porque me voy chao a mi I’m going to kill you

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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

a 5 sided solid has the numbers 1,2,3,4, and 5. what is the probability of rolling two five-sided solids and getting a sum of ei

kipiarov [429]

So this is going to be alot of writing to show my thinking but ill bold the answer.

1,1
1,2
1,3
1,4
1,5

2,1
2,2
2,3
2,4
2,5

3,1
3,2
3,3
3,4
3,5

4,1
4,2
4,3
4,4
4,5

5,1
5,2
5,3
5,4
5,5

next ill mark all the ones that equal 4 or 8 when added together, with an x

1,1
1,2
x1,3
1,4
1,5

2,1
x2,2
2,3
2,4
2,5

x3,1
3,2
3,3
3,4
x3,5

4,1
4,2
4,3
x4,4
4,5

5,1
5,2
x5,3
5,4
5,5

that is 6 (that equal 4 or 8) out of 25

so your ratio would be 6:19

6 0

3 years ago

X-5y=-15x−5y=−15x, minus, 5, y, equals, minus, 15 Complete the missing value in the solution to the equation. (-5,(−5,left paren

Y_Kistochka [10]

Answer:

<h2>The missing value is 2. The coordinate will be (-5, 2)</h2>

Step-by-step explanation:

The question is not properly written. Find the correct question below.

If x – 5y = -15 . Complete the missing value in the solution to the equation (-5, ____)

Let the coordinate of the variables be (x, y). Comparing the coordinates (x, y) with the given coordinate (-5, __), we will discover that x = -5. To get the y coordinate, we will substitute x = -5 into the given expression as shown;

If x – 5y = -15

-5 - 5y = -15

Adding 5 both sides

-5-5y+5 = -15+5

-5y = -10

Dividing both sides by -5;

-5y/-5 = -10/-5

y = 2

<u>Hence the missing value in the solution of the equation is 2</u>. The coordinate will be (-5, 2)

3 0

3 years ago

One pound of ground beef is $3.09. How much is 4.5 pounds? Round to the nearest hundredth, if necessary.