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Nikitich [7]
3 years ago
9

You have to find the ratio, rate

Mathematics
1 answer:
bonufazy [111]3 years ago
8 0

Answer:

8) 1 hour to drive 57 miles

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JulsSmile [24]

Answer:

2:10

Step-by-step explanation:

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3 years ago
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A manufacturing company plans to coat the entire exterior of a cylindrical shipping container with a 0.075-mm layer of special r
ycow [4]

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534.07cm² * 0.35=186.9245 dollar

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3 years ago
A(4, -1), B(3, 8), C(-1, 1); y = -2
nevsk [136]

Answer:

Point A: (4, -3)

Point B: (3, -12)

Point C: (-1, -5)

Step-by-step explanation:

Reflection of point A:

The x will stay the same and only the y will change.

Take the y-coordinate and subtract it from the reflection line while getting the absolute value..

|(-1) - (-2)| = |1| = 1

This means, point A is one unit down the line y = -2

(-2) - 1 = -3

The coordinate for A will now be (4, -3).

Reflection of point B:

Again, the x will stay the same and only the y will change.

Using the absolute value, get the y-coordinate and subtract it from the reflection line again..

|(8) - (-2)| = |10| = 10

This means, point A is five unit down the line y = -2

(-2) - 10 = -17

The coordinate for B will now be (3, -12).

Reflection of point C:

I think you get the gist of it.

|(1) - (-2)| = |3| = 3

(-2) - 3 = -5

The coordinate for C will now be (-1, -5).

I added an image down below of what the reflection looks like. Sometimes it helps to see things visually.

The red triangle is the original image and the blue triangle is the image after the reflection. The purple line is the reflection line y = -2

Hope this helped!

6 0
3 years ago
What is the opposite of factoring
ASHA 777 [7]

Answer:

simplifying

Step-by-step explanation:

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4 years ago
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(1+cos2x)/(1-cos2x) = cot^2x
sesenic [268]

We will turn the left side into the right side.

\dfrac{1 + \cos 2x}{1 - \cos2x} = \cot^2 x

Use the identity:

\cos 2x = \cos^2 x - \sin^2 x

\dfrac{1 + \cos^2 x - \sin^2 x}{1 - ( \cos^2 x - \sin^2 x)} = \cot^2 x

\dfrac{1 - \sin^2 x + \cos^2 x }{1 - \cos^2 x + \sin^2 x} = \cot^2 x

Now use the identity

\sin^2 x + \cos^2 x = 1 solved for sin^2 x and for cos^2 x.

\dfrac{\cos^2 x + \cos^2 x }{\sin^2 x + \sin^2 x} = \cot^2 x

\dfrac{2\cos^2 x}{2\sin^2 x} = \cot^2 x

\dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x

\cot^2 x = \cot^2 x


8 0
3 years ago
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