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3 years ago

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Russell and Aaron can build a shed in 8 hours when working together. Aaron works three times as fast as Russel. How long would i

t take Russel to build the shed if he were to work alone?

1 answer:

Whitepunk [10]3 years ago

4 0

Answer:

32 hours.

Step-by-step explanation:

Let x represent the time taken by Russell and Aaron to build the shed.

We have been given that Russell and Aaron can build a shed in 8 hours when working together. Aaron works three times as fast as Russel.

Russell's work rate would be $\frac{1}{x}$ .

Since Aaron works three times as fast as Russel, so Aaron's work rate would be $\frac{3}{x}$ .

Part of work done by in one hour would be $\frac{1}{8}$ .

We can represent our given information in an equation as:

$\frac{1}{x}+\frac{3}{x}=\frac{1}{8}$

Let us solve for x.

$\frac{1}{x}*8x+\frac{3}{x}*8x=\frac{1}{8}*8x$

$8+3*8=x$

$8+24=x$

$32=x$

Therefore, it will take Russell 32 hours to build the shed working alone.

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A woman earns $ 1,350 in interest from two accounts in one year. If she has three times as much invested at 7% as she does at 6%

VashaNatasha [74]

Answer:

The woman invested $15,000 at 7% interest rate and $5,000 at 6% interest rate.

Step-by-step explanation:

We are given the following in the question:

Let x be the interest earned from 7% interest rate and y be the interest earned from 6% interest rate.

The woman invested has three times as much invested at 7% as she does at 6%.

Thus, we can write the equation:

$x = 3y$

The total interest is $1,350.

Thus, we can write the equation:

$1350 = \dfrac{7x}{100} + \dfrac{6y}{100}\\\\7x + 6y = 135000$

Solving the two equations by substitution method:

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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!

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