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2 years ago

60 EXTRA POINTS!!!! I need help with writing an inequality for a graph. It is below. :)

Mathematics

1 answer:

ch4aika [34]2 years ago

8 0

Answer:

The answer to your question is: y ≥ - x + 6 and y ≥ x - 4

Step-by-step explanation:

Look of a pairs of points for each graph

Line 1 Line 2

A(0, 6) B (5, 1) C (7,3) D(5, 1)

m = (1- 6) / (5- 0) m = (1 - 3) / (5 - 7)

m = -5/5 = -1 m = -2 / -2 = 1

Find the line equation for each slope (y - y1) = m (x - x1)

(y - 1) = -1 (x - 5) (y - 1) = 1 (x - 5)

y - 1 = -x + 5 y -1 = x - 5

y = -x + 5 + 1 y = x - 5 + 1

y = -x + 6 y = x - 4

Inequality

y ≥ - x + 6 y ≥ x - 4

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Ebony earned scores of 72 70, 81, 99, and 88 on five math tests. Determine her

Minchanka [31]

Let it be x

$\boxed{\sf Average=\dfrac{Sum\:of\:terms}{No\:of\:terms}}$

$\\ \sf\longmapsto \dfrac{72+70+81+99+88+x}{6}=83$

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2 years ago

What shape is a rhombus related to and why

Ksju [112]

A rhombus is related to a “Parallelogram”

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2 years ago

Solve the equation.<br><br> -1 2/7 ÷ 1 13/14 =

sergeinik [125]

Answer:

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Step-by-step explanation:

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5 0

2 years ago

Read 2 more answers

Is the given number a solution of the equation?<br> 8=2+3; 10

kykrilka [37]

Answer:

No.

Step-by-step explanation:

8 doesn't equal 0 on the very right, which means it's not a solution to the equation. So therefore, it is false based on the equation.

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2 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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2 years ago