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frez [133]
3 years ago
14

You and a friend missed the bus so you must walk home from school together. At the park, you part ways.

Mathematics
2 answers:
aniked [119]3 years ago
5 0

Answer:

10 kilometers

Step-by-step explanation:

4 kilometers north plus 6 kilometers east makes 10 kilometers. This is the right answer.

Cerrena [4.2K]3 years ago
5 0

Answer:

10

Step-by-step explanation:

because if you part ways after walking the same distance and he walks 4 kilometers and you walk 6 well 4+6=10.

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What is the length of DF? Round to the nearest hundredth.
OlgaM077 [116]

Answer:

7.06

Step-by-step explanation:

This triangle can be solved a couple of ways. In the end, they amount to the same thing.

1) The area is ...

A = 1/2bh = 1/2(8)(15) = 60 . . . using DG as the base

Using GE as the base, the height (DF) is ...

A = (1/2)(17)(DF)

2(60)/17 = DF = 120/17

DF ≈ 7.06

__

2) Using similar triangles, we can find the ratio of the long side to the hypotenuse as ...

(long side)/(hypotenuse) = DE/GE = DF/DG

DF = DG(DE/GE) = 8(15/17) = 120/17

DF ≈ 7.06

3 0
3 years ago
−2(2x + 5) − 3 = −3(x − 1)
nasty-shy [4]

Answer:

x=−16

Step-by-step explanation:

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

distribute -2 into (2x+5) and -3 into (x-1)

-4x-10-3=-3x+3

combine like terms

-4x-13=-3x+3

add -4x on both sides

-13=-3x+4x+3

combine like terms

-13=x+3

subtract 3 on both sides

-16=x

hope this helps

4 0
3 years ago
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maks197457 [2]

Answer:

3.14

Step-by-step explanation:

Formula for the area of a circle:

A = πr^2

put in the radius into the equation,

A = π(1)^2

8 0
3 years ago
The last one is $30 which one ?
12345 [234]

Answer:

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Step-by-step explanation:

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6 0
3 years ago
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An angle t is drawn from the center of the unit circle. Find a formula in terms of t for the straight line distance d between th
Doss [256]

Answer:

d = \sqrt{2}\cdot \sqrt {1-\cos t - \sin t}

Step-by-step explanation:

Let suppose that one of the radii meets the circle at the point (1,0). The straight line distance formula is:

d = \sqrt{(\cos t - 1)^{2}+(\sin t - 1)^{2}}

d = \sqrt{(\cos^{2}t - 2\cdot \cos t + 1)+(\sin^{2}t - 2\cdot \sin t + 1)}

d = \sqrt{2-2\cdot (\cos t + \sin t )}

d = \sqrt{2}\cdot \sqrt {1-\cos t - \sin t}

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