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2 years ago

7

the temperature on the ground during a plane's takeoff was 7° F. At 38,000 feet in the air, the temperature outside the plane wa

s -36° F. Find the difference between these two temperatures, with the number only.

2 answers:

arlik [135]2 years ago

7 0

Just for point have a good day

iragen [17]2 years ago

3 0

43 degrees farenheit

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What is the square root of 9?

Elza [17]

The square root of nine is three.

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2 years ago

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Andre wants to save $40 to buy a gift for his dad. Andre’s neighbor will pay him weekly to mow the lawn, but Andre always gives

Dmitry_Shevchenko [17]

(I could answer more confidently if I had more context to the question)

I believe the answer to the question is Andre's neighbor pays Andre 10 dollars a week to mow his lawn.

Hope this helps, if not, comment below please!!

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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

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

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3 years ago

Given an int variable n that has already been declared and initialized to a positive value, and another int variable j that has

Inessa05 [86]

Answer:

for(j = n; j > 0; j--)

System.out.print(\"*\");

Step-by-step explanation:

-A for loop is a repetition control structure.

-It allows the efficiency to write a loop that would otherwise be written a couple of times.

-The output is a single line comprising n asterisks,

-The number of asterisks printed will be equivalent to the n-value declared.

7 0

2 years ago

Graph<img src="https://tex.z-dn.net/?f=fx%3D%20%7Cx%2B1%7C%20%2B2" id="TexFormula1" title="fx= |x+1| +2" alt="fx= |x+1| +2" alig

AURORKA [14]

Answer:

The graph is shown in the following image

Step-by-step explanation:

The parent function of $f(x) = | x + 1 | +2$ is $y = | x |$

The graph of $y = | x |$ is the union of two lines:

$y = x$ for $x\geq0$

$y = -x$ for $x$

The vertex of this graph is in point (0,0)

<em>The graphic is shown in the attached image</em>.

Then, if we do $y = f(x + 1)$ to the parent function we get:

$y = | x + 1 |$

This transformation shifts the graph of $y = |x|$ 1 unit to the left

Then, if we make the transformation $y = f(x) + 2$ to the function $y = | x + 1 |$ we obtain:

$y = | x + 1 | +2$

This transformation displaces the graph of the function $y = | x + 1 |$ 2 units up on the y axis.

So, if the vertex of the main function $y = | x |$ was at point (0,0), the vertex of the function $f(x) = |x + 1| +2$ has its vertex in the point (-1, 2).

<em>The graph is shown in the following image</em>

5 0

3 years ago