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3 years ago

Help me plzzzz!!!! Graph: y-3=1/2(x+2)

Mathematics

1 answer:

Zolol [24]3 years ago

8 0

Answer:

Step-by-step explanation:

First you need to find the Slope Intercept form or find the point and slope.

The point is: (-2,3)

and the slope is: 1/2

y - 3 = 1/2(x + 2)

y - 3 = 1/2x + 1

y = 1/2x + 1 + 3

y = 1/2x + 4

Slope intercept form: y = 1/2x + 4

Graph is show below ---

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Rewrite the expression 7x+21=

Dima020 [189]

Minus the 21 from the other side of the equation making it 7x=-21 then u divide the 7 from the x to the -21 resulting in x=-3

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Which of the following expressions represents the expanded version of the expression

maxonik [38]

Answer:

4th option

Step-by-step explanation:

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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

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

Find the <br> (D)<br> (R)<br> &<br> (A)

omeli [17]

<h3>Answer: Choice C</h3>

domain = (-infinity, infinity)
range = (-infinity, 0)
horizontal asymptote is y = 0

========================================================

Explanation:

Since no division by zero errors are possible, and other domain restricting events are possible, we can plug in any x value we want. This means the domain is the set of all real numbers. Representing this in interval notation would be (-infinity, infinity).

The range is the set of negative real numbers, which when written in interval notation would be (-infinity, 0). This is because y = 5^x has a range of positive real numbers, and it flips when we negate the 5^x term. The graph of y = -5^x extends forever downward, and the upper limit is y = 0.

It never reaches y = 0 itself, so this is the horizontal asymptote. Think of it like an electric fence you can get closer to but can't touch.

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3 years ago