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3 years ago

14

Employe work 8 hours on 2 days 6 hours on 1 day and 4 hours on 2 days what’s the average number

1 answer:

Vladimir [108]3 years ago

3 0

The average number that the employee works is 6 hours

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Use an array or an area model to illustrate how to solve this equation. Then, explain each part of the

valentina_108 [34]

Answer:

870

Step-by-step explanation:

Array model shown below along with description (at the bottom).

3 0

4 years ago

Solve for p.<br><br> 5=4p−13<br><br><br> p=72<br><br><br> p = 4<br><br><br> p = 7<br><br><br> p = 64

Natali [406]

Answer:

I think it's 4

Step-by-step explanation:

7 0

2 years ago

Three fourths of a number is 1/2 find the number in lowest terms

NNADVOKAT [17]

Answer:

To thirds if I'm not wrong

Step-by-step explanation:

three fourths meaning 3/4 of a number is 1/2

so:

3/4 of number is 1/2

1/4 of number = number ÷ 4

3/4 of number = (number ÷ 4) × 3

(number ÷ 4) × 3 = 1/2

<divide both sides by 3>

number ÷ 4 = 1/6

<multiply both sides by 4>

number = 4/6

<show in lowest term>

number = 2/3

the number is 2/3

5 0

4 years ago

Find the value of <br> g(−2) if g(x)=x3−x

Angelina_Jolie [31]

$g(x)=x^3-x\\g(-2)=(-2)^3-(-2)\\g(-2)=-8+2\\g(-2)=-6$

Therefore, g(-2) = -6

Basically substitute x = -2 in the equation.

4 0

3 years ago

Differentiate with respect to x and simplify your answer. Show all the appropriate steps? 1.e^-2xlog(ln x)^3 2.e^-2x(log(ln x))^

serious [3.7K]

(1) I assume "log" on its own refers to the base-10 logarithm.

$\left(e^{-2x}\log(\ln x)^3\right)'=\left(e^{-2x}\right)'\log(\ln x)^3+e^{-2x}\left(\log(\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{e^{-2x}}{\ln10(\ln x)^3}\left((\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10(\ln x)^3}\left(\ln x\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10\,x(\ln x)^3}$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}}{\ln10\,x\ln x}$

Note that writing $\log(\ln x)^3=3\log(\ln x)$ is one way to avoid using the power rule.

(2)

$\left(e^{-2x}(\log(\ln x))^3\right)'=(e^{-2x})'(\log(\ln x))^3+e^{-2x}\left(\log(\ln x))^3\right)'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2(\log(\ln x))'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2\dfrac{(\ln x)'}{\ln10\,\ln x}$

$=-2e^{-2x}(\log(\ln x))^3+\dfrac{3e^{-2x}(\log(\ln x))^2}{\ln10\,x\ln x}$

(3)

$\left(\sin(xe^x)^3\right)'=\left(\sin(x^3e^{3x})\right)'=\cos(x^3e^{3x}(x^3e^{3x})'$

$=\cos(x^3e^{3x})((x^3)'e^{3x}+x^3(e^{3x})')$

$=\cos(x^3e^{3x})(3x^2e^{3x}+3x^3e^{3x})$

$=3x^2e^{3x}(1+x)\cos(x^3e^{3x})$

(4)

$\left(\sin^3(xe^x)\right)'=3\sin^2(xe^x)\left(\sin(xe^x)\right)'$

$=3\sin^2(xe^x)\cos(xe^x)(xe^x)'$

$=3\sin^2(xe^x)\cos(xe^x)(x'e^x+x(e^x)')$

$=3\sin^2(xe^x)\cos(xe^x)(e^x+xe^x)$

$=3e^x(1+x)\sin^2(xe^x)\cos(xe^x)$

(5) Use implicit differentiation here.

$(\ln(xy))'=(e^{2y})'$

$\dfrac{(xy)'}{xy}=2e^{2y}y'$

$\dfrac{x'y+xy'}{xy}=2e^{2y}y'$

$y+xy'=2xye^{2y}y'$

$y=(2xye^{2y}-x)y'$

$y'=\dfrac y{2xye^{2y}-x}$

8 0

3 years ago