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poizon [28]
3 years ago
11

Got 1+7 Do this problem to find out.

Mathematics
1 answer:
lidiya [134]3 years ago
4 0

tueiwgeirgrurhrhe8fhtu3i

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What is the slope of the line that passes through the points (-6, 8) and (-9, 7)? Write your answer in simplest form.
yanalaym [24]

Answer:

m=1/3

equation y=1/3x+10

Step-by-step explanation:

Find the rise and run.

1. -6--9=run=3

2. 8-7=rise=1

6 0
2 years ago
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What is the average rate of change of f(x), represented by the table of values, over the interval [-3, 2]?
maksim [4K]
You would select the points with the x-values -3 and 2 and use the slope formula:
(-3, -36) and (2,4)
m= \frac{y_2-y_1}{x_2-x_1} ← slope formula
x1 = -3 y1 = -36 and x2 = 2 y2 = 4

m= \frac{4--36}{2--3} = \frac{40}{5} =8
3 0
3 years ago
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At a farm the ratio of cows to horses was 9:2. If there were 72 cows at the farm how many horses were there?
AlladinOne [14]
16 horses. Since 9 x 8 = 72, then 2 x 8 = 16
7 0
3 years ago
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A 41-foot ladder is placed against a vertical wall of an apartment building. The base of the ladder is 9 feet from the base of t
amid [387]

The ladder reaches 40 feet up the wall.

41 squared (1681) minus 9 squared (81) is 1600. 40 is the square root of 1600.

6 0
3 years ago
Use a(t) = −9.8 meters per second per second as the acceleration due to gravity. (Neglect air resistance.) A canyon is 2100 mete
Umnica [9.8K]

Answer:

t \approx 20.7\,s

Step-by-step explanation:

Given that rock experiments a constant acceleration, position function can be obtained by integrating twice:

v(t) = v_{o} - a\cdot t

s(t) = s_{o} + v_{o}\cdot t -\frac{a}{2}\cdot t^{2}

The initial conditions of the rock are, respectively:

s_{o} = 2100\,m

v_{o} = 0\,\frac{m}{s}

Position of the rock as a function of time is:

s(t) = 2100\,m -(4.9\,\frac{m}{s^{2}})\cdot t^{2}

The time taken for the rock to hit the canyon floor is:

0\,m = 2100\,m - (4.9\,\frac{m}{s^{2}} )\cdot t^{2}

2100\,m = (4.9\,\frac{m}{s^{2}} )\cdot t^{2}

t = \sqrt{\frac{2100\,m}{4.9\,\frac{m}{s^{2}} } }

t \approx 20.7\,s

8 0
3 years ago
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