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2 years ago

Please someone Help me 40 points the answer shown Down Below may not be the Correct one

Mathematics

2 answers:

FrozenT [24]2 years ago

3 0

I think A or C hope this helps

Lynna [10]2 years ago

3 0

Answer:

The answer is A

Step-by-step explanation:

Boi gimme dat Brainlyyyyy

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Find the volume of the solid. Round to the nearest tenth if necessary.

Vikki [24]

$V= \frac{4}{3} \pi r^{3} or V=4.1888 r^{3}$

r=4 $r^{3}$ =64*4.1888=268.083

V = 268.083 $mm^{3}$ so go with A.

7 0

2 years ago

What is the equivalent of 3/6

Maslowich

$\frac{3}{6}^{/3}= \frac{1}{2}$

3 0

2 years ago

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(2x -2)^-3<br>can anyone help me with this?

zzz [600]

Answer: See below

Step-by-step explanation:

When you see that something is to the power of a negative number, it is another way of saying 1 over.

$(2x-2)^-^3=\frac{1}{(2x-2)^3} =\frac{1}{(2x-2)(4x^2-8x+4)} =\frac{1}{8x^3-8x^2+8x-8}$

4 0

3 years ago

Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

2 years ago

Mr. Burn's class is taking a field trip to the planetarium. The trip will cost $27 for each

marysya [2.9K]

Answer:

$486

Step-by-step explanation:

all you do is multiply 27 and 18. i feel like using compatible numbers bc the numbers you have will give you your answer.

7 0

2 years ago

Read 2 more answers