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Rasek [7]
3 years ago
15

How many millimeters are there in 2 kilometers

Mathematics
2 answers:
Vesnalui [34]3 years ago
8 0
There are 1000 millimeters in 1 kilometer

1000mm = 1 km 
2000mm = 2km
Ugo [173]3 years ago
3 0
There are 2000000 millimeters in 2 kilometers
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ect the best answer for the question. 3. Rewrite the equation x + y + 3 = –2x + 4y + 18 into standard form (ax + by = c). A. 3x
trapecia [35]

Answer:

B

Step-by-step explanation:

To rewrite the equation x + y + 3 = -2x + 4y + 18, first combine like terms.

x + y + 3 = -2x + 4y + 18

3x - 3y + 3 = 18

3x - 3y = 15

This is the standard form.

It can also be simplified by dividing each term by 3 to x - y = 5.

5 0
3 years ago
Determine the equation of the circle graphed below.
erica [24]

Answer: (-7,3)

Step-by-step explanation:

3 0
3 years ago
Find the area of the composite shape. Use 3.14 as pi.<br> TENTH! Use your formula sheet!
murzikaleks [220]
59.6 is the answer rounded to the nearest tenth
8 0
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
5 0
3 years ago
Use number properties to simplify the following expression. -5 + (5 + 3)
bija089 [108]
Number Properties: PEMDAS

Parentheses, Exponents, Multiplication, Division, Addition, & Subtraction


- 5 + (5 + 3)

1. Parentheses
- 5 + (7) \\
There Are No Exponents, Multiplication, Division, Or Addition

2. Subtraction
2
7 0
3 years ago
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