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babunello [35]
3 years ago
7

Help! Match each linear equation with its graph

Mathematics
1 answer:
Bond [772]3 years ago
6 0

Answer:

y = -1/4 x+4 is blue

y = 1/2 x+4 is green

y = 3x is red

y = -1/2x is purple

y = 3 is black

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PLS HELP ME GIVING 43 POINTS I GIVE BRANLIEST PLS HELP ME FOR THIS TEST ITS MATH SUPER EZ
Naya [18.7K]

Answer:lydra=21,horatio=17

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What expression is equivalent to the expression (4x+12) + 7x?
pochemuha

Answer:

11x + 12

Step-by-step explanation:

( 4x + 12 ) + 7x

Remove parentheses.

= 4x + 12 + 7x

Group like terms.

= 4x + 7x + 12

Add similar elements: 4x + 7x = 11x

= 11x + 12

8 0
3 years ago
Read 2 more answers
16 points<br> 1) Write the following in standard form (answer). *<br> 4^5
m_a_m_a [10]

Answer:

1024

Step-by-step explanation:

You could put 4^5 into your calculator. It’s 4*4*4*4*4= 1024

3 0
3 years ago
Need help asap for test
alex41 [277]

Answer:

a=4 and b=-1

Step-by-step explanation:

2(4)+3(-1)= 5

6 0
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
5 0
3 years ago
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