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ankoles [38]
3 years ago
9

Simplify 8 + 2(10 – r). 

Mathematics
2 answers:
r-ruslan [8.4K]3 years ago
4 0
8+20-2r which is equal to 28-2r so it is (B)
balu736 [363]3 years ago
3 0

Hi!

8 + 2(10 - r) = 8 + 20 - 2r = 28 - 2r

The correct answer is B. 28 - 2r

Hope this helps!

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The temperature in frostville was at 50*F at midnight,and it dropped at a constant rate for the next 10 hours.
bazaltina [42]

After 10 hours the temperature is shown on the graph as 30 degrees.

50 degrees - 30 degrees = 20 degrees.

The temperature dropped 20 degrees in 10 hours.

Divide the change by the time:

20 degrees / 10 hours = 2 degrees per hour.

Because the temperature dropped, the change would be negative.

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3 years ago
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3 years ago
A ball is thrown into the air with a velocity of 34 ft/s. It's height in feet after t seconds is given by y=34t-26(t)^2. Find th
kkurt [141]

Answer:

The average velocity of the ball at the given time interval is -122.3 ft/s

Step-by-step explanation:

Given;

velocity of the ball, v = 34 ft/s

height of the ball, y = 34t - 26t²

initial time, t₀ = 3 seconds

final time, t = 3 + 0.01 = 3.01 seconds

At t = 3 s

y(3) = 34(3) - 26(3)² = -132

The average velocity of the ball in ft/s is given as;

V_{avg} = \frac{y(3.01)-y(3)}{3.01 -3}\\\\V_{avg} = \frac{34(3.01)-26(3.01)^2-y(3)}{3.01 -3}\\\\V_{avg} = \frac{-133.223-y(3)}{0.01}\\\\V_{avg} = \frac{-133.223-(-132)}{0.01}\\\\V_{avg} =\frac{-1.223}{0.01}\\\\V_{avg} = -122.3 \ ft/s

Therefore, the average velocity of the ball at the given time interval is -122.3 ft/s

8 0
3 years ago
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