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2 years ago

The putting together of two groups is called

Mathematics

1 answer:

den301095 [7]2 years ago

4 0

It is called a bonding kilyodhuifguegdhegufguygfgerugued

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Evaluate the expression for the given value of the variable |b|+b³;b=-2

erastova [34]

|b| + b^3; b = -2

Substitute -2 for b.

|(-2)| + (-2)^3

Solve exponents first.

|(-2)| + (-8)

|(-2)| is the same as positive 2 since it is the absolute value.

Rewrite the expression.

2 + (-8)

Solve.

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2 years ago

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11Alexandr11 [23.1K]

the answer is infact c!

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1 year ago

Help me with geometry?

Sliva [168]

C is a total of 90 degrees

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2 years ago

What is the surface area of the figure? Use 3.14 for π....PLEASE HELP....NEED HELP ASAP

raketka [301]

Answer:

Step-by-step explanation:

Always use your imaginary scissor:

Cut the above and below circle and the cylinder role it down-ways like a carpet and you will end up with two circles and a big rectangular.

Surface circle: 7 * 7 * 3,14 = 153.86

Surface carpet: 20 * circle exterior (or the length of the circle)= 20 * 2* 3.14 * 7

Results:

153, 86 * 2 (because we cut 2 circles from the cylinder) + 132,6 = 439.6

Answer: A

3 0

2 years ago

Read 2 more answers

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

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

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1 year ago