What is the domain of the relation shown<br> on the graph?

How much was earned after 5 hours of work?

Ugo [173]

62.5

37.5 - 25 = 13.5
3*5= 62.5

4 0

3 years ago

Read 2 more answers

Use synthetic division to find (x^2+3x–4)÷(x+4)

Jobisdone [24]

Answer:

X^2-x- 8/x+4

Step-by-step explanation:

X+4 becomes a negative 4

Then multiply -4 by the first number which is 1 so -4×1 is -4. Then subtract -4 by the second number 3. This gives you -1 so multiply that by 4 which is -4. Then subtract 4 from the third number -4. So that'll give you -8. Then add an x-value to every number except the very last number as that number will be you remainder.

7 0

3 years ago

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A machine needs 2 of its gears replaced. You find 2 gears, one manufactured 5 years ago and the other manufactured 10 years ago.

andrey2020 [161]

These are independent events so
P(A and B) = P(A) * P(B)

Therefore
P(Both gears fail) = 0.05 * 0.08 = 0.004 or 0.4%

3 0

3 years ago

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$

5 0

3 years ago

Ramla bought a sheet of 100 stamps from the post office for $39. what was the price of each stamp?

erica [24]

The price of each stamp was $2.56 each because 100 divided by $39 is $2.56.

Hope that helped:)

5 0

3 years ago

What is the domain of the relation shown on the graph?