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2 years ago

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The manufacturer is considering production quantities of 40,000 units or 80,000 units. For the 40,000 unit plan, assume 95% of p

roduct will be sold and 5% will be salvaged; for the 80,000 unit plan, assume 80% of product will be sold and 20% will be salvaged.

1 answer:

DIA [1.3K]2 years ago

7 0

Bro got school in summer

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PLEASE SOMEONE HELP ME!!! I WILL GIVE BRAINLIEST TO CORRECT ANSWER!!! PLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS

taurus [48]

8.8 squared + 9.5 squared = square root 12.94 rounded 12.1

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3 years ago

In the diagram, point D divides line segment AB in the ratio of 5:3. If line segment AC is vertical and line segment CD is horiz

KIM [24]

Answer:

D

Step-by-step explanation:

First, find the coordinates of pointD if A(2,-6), B(10,2) and point D divides line segment AB in the ratio of 5:3.

If point D divides the segment AB in the ratio m:n, then

$D\left(\dfrac{nx_A+mx_B}{m+n},\dfrac{ny_A+my_B}{m+n}\right)$

So

$D\left(\dfrac{3\cdot 2+5\cdot 10}{5+3},\dfrac{3\cdot (-6)+5\cdot 2B}{5+3}\right)=D(7,-1)$

Point C has the x-coordinate the same as point A and the y-coordinate the same as point D.

Thus, C(2,-1)

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3 years ago

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How to find the surface area and the volume of this prism

kotegsom [21]

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3 years ago

What is the formula to the area of a triangle?

Lorico [155]

Answer:

(Base x height) x 1/2

Step-by-step explanation:

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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!

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3 years ago