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3 years ago

8

Mikaela placed a frame around a print that measures 10 inches by 10 inches. The area of just the frame itself is 69 square inche

s. What is the width of the frame?

1 answer:

olchik [2.2K]3 years ago

3 0

Answer:

1.5 in

Step-by-step explanation:

Let x be the width of the frame.

Side of print=10 in

Area of frame=69 square in

We have to find the width of the frame.

Side of frame=10+x+x=10+2x

Area of square= $(side)^2$

By using the formula

Area of print= $10\times 10=100in^2$

Area of frame with print= $(10+2x)^2$

Area of frame=Area of frame with print-Area of print

$69=(10+2x)^2-100$

$(10+2x)^2=69+100=169$

$(10+2x)=\sqrt{169}=\pm 13$

$10+2x=13$

Because Side is always positive.

$2x=13-10=3$

$x=\frac{3}{2}=1.5$

Hence, width of frame=1.5 in

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3 years ago

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3 years ago

Read 2 more answers

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 $y=2$ :

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So the area of $R$ is given by

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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

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

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

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3 years ago