One printing machine printed 700 books in 40 hours this week. A second printing machine printed for

Mathematics

1 answer:

Mamont248 [21]3 years ago

7 0

Answer:

875 books

Step-by-step explanation:

1st machine 700 books/40 hours = 17.5 books/hr

2nd machine printed 2x or 35 books/hr

25 hours x 35 books/hr = 875 books

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HELP ME I DON'T UNDERSTAND THIS :(

andriy [413]

Answer:

Step-by-step explanation:

Since the figures are similar then the ratios of corresponding sides are equal, that is

$\frac{AD}{WZ}$ = $\frac{CD}{YZ}$ , substitute values

$\frac{10}{46}$ = $\frac{5}{YZ}$ ( cross- multiply )

10YZ = 230 ( divide both sides by 10 )

YZ = 23 → c

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

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