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Zielflug [23.3K]
3 years ago
7

In 14 days the daily high temperature in cedar hills dropped by 16.8 degrees Fahrenheit what is the average daily rate of change

in the high temperature in degrees Fahrenheit
Mathematics
1 answer:
Montano1993 [528]3 years ago
6 0

Answer:

1.2°F

Step-by-step explanation:

Given that:

The drop in temperature in degree Fahrenheit over a period of 14 days = 16.8

The average daily rate of change in degwrr Fahrenheit ;

Total drop in temperature / number of days

= 16.8°F / 14

= 1.2°F

Hence, the average daily drop in temperature is 1.2°F

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Answer:

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Step-by-step explanation:

Assuming from the cut off picture, that we need to find the slope of ...

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m is the slope and b is the y-intercept. Anything number that is in place of m, is the slope; and any nother in place of b is the y-intercept.

In that case, 2.25 is the slope and 3 is the y-intercept

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3 years ago
The height of a football during a punt is modeled by h=-16t^2+60t+3. If the football hits the ground, how long did it stay in th
weqwewe [10]

"how long..." is asking for time (t).  "The amount of time spent in the air" is the time from when the ball was kicked (0 seconds) to the time it landed on the ground.  Need to find the x-intercepts (one will be negative which will be invalid). You can do this by factoring ... <em>or by using the quadratic formula.</em> With the equation you provided, it is not factorable, so you must use the quadratic formula.

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   a=-16   b=60  c=3

t = \frac{-b +/- \sqrt{b^{2}-4ac } }{2a}

t = \frac{-60 +/- \sqrt{60^{2}-4(-16)(3) } }{2(-16)}

t = \frac{-60 +/- \sqrt{3600 + 192} }{-32}

t = \frac{-60 +/- \sqrt{3792} }{-32}

t = \frac{-60 +/- 61.6}{-32}

t = \frac{-60 + 61.6}{-32}   or   t = \frac{-60 - 61.6}{-32}

t = \frac{1.6}{-32}        or     t = \frac{-121.6}{-32}

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Answer: 3.8 seconds


     

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Talja [164]
30 percent will be filled I am positive that is it
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The area of a circular fountain is 121pi square feet. What is the diameter of the fountain?
Alla [95]

Answer:

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Step-by-step explanation:

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7 0
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