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enyata [817]
3 years ago
14

Graph the solution to the following inequality.

Mathematics
1 answer:
e-lub [12.9K]3 years ago
5 0

Explanation= You basically do x=6 on the graph then you shade in the right side because that represents that x is GREATER THAN OR EQUAL to 6.

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Look at the picture
ValentinkaMS [17]

Answer: 54

Step-by-step explanation:

3 times the square of -3 plus -5 times -3 plus 12

3 times 9 + 15 + 12

27+15+12=

54

7 0
3 years ago
A company has two separate offices. One office has a color printer and the other office has a black and white
Vitek1552 [10]

Answer:

the larger of two printers being used to print the payroll for a major corporation requres 40 min. to print the payroll.So the smaller printer's payroll printing rate is  

1 payroll per x min or  or  

>>...After both printers have been operating for 10 min, the larger printer malfunctions...<<

So the fraction of the payroll that the larger printer did in the 10 minutes

is  = payroll.

Step-by-step explanation:

7 0
3 years ago
Use x=5, equals, 5 to identify the value of each expression.
Lesechka [4]

Answer:

The answer to the first question is 625

The answer to the second question is 1

The answer to the third question is 1

4 0
3 years ago
Read 2 more answers
Dan has 3 ⅔ cups of milk. If he used ¼ of the milk for his breakfast, how much milk does Dan have left?
riadik2000 [5.3K]

There would be 3 2/12 left.

2/3 x 1/4 = 2/12

Hope this helps!

Sorryyyy :C

6 0
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:

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
5 0
3 years ago
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