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Neporo4naja [7]
3 years ago
10

a football is kicked into the air with an initial upward velocity of 82 feet per second its height H in feet is recorded at vari

ous second in the table below write an equation for the curve of best fit than estimate the height of the football after 5 Seconds​

Mathematics
1 answer:
slamgirl [31]3 years ago
6 0

We have been given that a football is kicked into the air with an initial upward velocity of 82 feet per second. The height (H) of ball in feet is recorded at various second in the table. We are asked to write an equation that represents the given scenario.

We know that equation of height of an upward moving object is h(t)=-\frac{1}{2}gt^2+V_0t+h_0, where,

g = Acceleration due to gravity,

V_0 = Initial velocity,

h_0 = Initial height.

V_0=82 and h_0=0. We know that acceleration due to gravity is a constant and its value is 32 feet square per sec.

Upon substituting these values, we will get:

h(t)=-\frac{1}{2}(32)t^2+82t+0

h(t)=-16t^2+82t

Therefore, the equation h(t)=-16t^2+82 represents the height of the football after t seconds.

To find the height of the football after 5 seconds, we will substitute t=5 in our equation as:

h(5)=-16(5)^2+82(5)

h(5)=-16(25)+410

h(5)=-400+410

h(5)=10

Therefore, the height of the football after 5 seconds would be 10 feet.

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4 years ago
Simplify this expression.
Bond [772]

The correct expression to simplify is:

1+4.25n + 3/2p -3+(-2p)+5/4n

Answer:

5.5n-1/2p-2

Explanation:

<u>1. Given expression</u>:

1+4.25n + 3/2p -3+(-2p)+5/4n

<u>2. Group like terms</u>:

(1-3)+(4.25n+5/4n)+(3/2p-2p)

<u>3. Simplify</u>:

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4 years ago
A test taker gets 70 on 1st exam, 80 on 2nd exam, 2/3 of 4/5 of his 2nd exam on his 3rd test. If the professor gives 5 points ex
Serga [27]
<h3>Answer:  127</h3>

=================================================

Explanation:

The phrasing "2/3 of 4/5 of his 2nd exam on his 3rd test" is a bit clunky in my opinion. It seems more complicated than it has to be.

The student got 80 on the second exam. 4/5 of this is (4/5)*80 = 0.8*80 = 64. Then we take 2/3 of this to get (2/3)*64 = 42.667 approximately. If we assume only whole number scores are given, then this would round to 43.

Let x be the score on the fourth exam. Since 5 points of extra credit are given, the student actually got x+5 points on this exam.

So we have these scores

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  • second exam = 80
  • third exam = 43
  • fourth exam = x+5

Adding up these scores and dividing by 4 will get us the average

(sum of scores)/(number of scores) = average

(70+80+43+x+5)/4 = 80

(x+198)/4 = 80

x+198 = 4*80

x+198 = 320

x = 320 - 198

x = 122

So the student got a score of x+5 = 122+5 = 127 on the fourth exam.

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3 years ago
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Step-by-step explanation:

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