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Degger [83]
3 years ago
14

What is the slope of the line that passes through the points (0, -7) and (-4, 3)?

Mathematics
1 answer:
AleksAgata [21]3 years ago
8 0
B. -5/2 is the correct answer
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Question 8 help please, please provide proper explanation! BIG POINTS
Likurg_2 [28]

Answer:

176 cm

Step-by-step explanation:

45 degrees is what fraction of the circle

45/360  = 1/8

So multiply the arc length by 8 to get the circumference of the entire circle

22* 8 =176 cm

3 0
4 years ago
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James is working on a social studies project using a map with a scale of 60 miles equals 1 inches.
natima [27]

Answer:

3 miles =3 inch

200miles =200 inch

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8 0
3 years ago
one pipe fills a store pool in 12 hours. A second pipe fills the same pool in 6 hours. When a third pipe is added and all three
Alina [70]

Answer:

x  = 12  hours

Step-by-step explanation:

If one pipe fills store pool in 12 hours   in 1 hour will fill 1/12 of the pool

If one pipe fills a store pool in 6 hours in 1 hour will fill 1/6 of the pool

Let call " x " number of hours needed by the third pipe to fills the pool

then in 1 hour  will fill 1/x

The three pipes working together in 1 hour will fill

1/12  +  1/6  + 1/x

Now we know that the three pipes take 3 hours to fill the pool then in 1 hour three pipes will fill 1/3 of the pool, therefore

1/12  +  1/6  + 1/x  = 1/3

( x + 2*x + 12) /12*x  = 1/3   ⇒ ( 3*x +12 )/ 12*x = 1/3

9*x  + 36  = 12*x

- 3*x  = - 36

x  = 36/3

x  = 12  hours

5 0
3 years ago
What is 1/100 as a percent?
KonstantinChe [14]
Simple it is 1% because it is 1 out of 100 parts equaling 1%

7 0
3 years ago
Can anyone just explain the steps to this problem I keep messing up and I need some guidance. Thank you!
timama [110]

Answer:

(B) 1

Step-by-step explanation:

g(x) = ∫₂ˣ f(t) dt

Use second fundamental theorem of calculus to find g'(x).

g'(x) = f(x) d(x)/dx − f(2) d(2)/dx

g'(x) = f(x)

g(x) is a maximum when g'(x) = 0.

0 = f(x)

x = 3

So the maximum value of g(x) is:

g(3) = ∫₂³ f(t) dt

g(3) = ½ (2) (1)

g(3) = 1

7 0
4 years ago
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