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3 years ago

How do i find the volume of a composite figure?

1 answer:

5 0

-- Break up the composite figure into pieces
that you know how to find the volume of.

-- Find the volume of each piece.

-- Add up the volumes of all the pieces.

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I really need help plz help me

Phantasy [73]

Answer:

Step-by-step explanation:

This is actually really easy you just have to subtract 36 from 100

Hope this helps also please mark brainliest

6 0

3 years ago

Pam read 126 pages of her summer reading book in 3 hours. Zack read 180 pages of his summer reading book in 4 hours. If they con

LiRa [457]

Zack reads faster so he will finish first
Pam= 126/3 = 42/1
Zack= 180/4 =45/1
Now multiply the unit rates by 5 each
Pam= 210/5
Zack= 225/5

8 0

3 years ago

Read 2 more answers

Line segments AB and CD intersect at point E. Ray EF extends from point E.

Tom [10]

Answer:

x=29 CEF=103 BEF=58

Step-by-step explanation:

161=3x+16+2x

5x=145

x=29

7 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

3 years ago

Given the function f(x) = 3x3 + 7x2 + 5x, find f(-3).

Julli [10]

Answer:

the ans is b -33

Step-by-step explanation:

by putting the -3 in the place of x and adding,subtracting

4 0

3 years ago