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madam [21]
2 years ago
13

It’s the start of a new football season and my favorite team, the WACO Hawks, is looking good! They won their first

Mathematics
1 answer:
Evgesh-ka [11]2 years ago
6 0

Vernon Variable ran 43 yards, Reggie Root ran 114 yards, Larry Linear ran 54 yards, and Quentin ran 18 yards.

<h3>How to solve Linear Equations Word Problems?</h3>

We are told that the total number of yards the WACO Hawks ran is 229. Thus, we can say that: W = 229

Now, we know that Vernon Variable gained many yards. Thus, the amount of yards will be indicated by "V"

Now, Reggie Root ran for 15 less than three times as many yards as Vernon did. Thus, distance Reggie root ran is represented by;

R = 3V-15

Larry Linear picked up 11 more yards than Vernon’s total. Thus, total yards by Larry Linear is: L = V + 11

Quentin Quadratic ran a total of 18 yards. Thus; Q=18.

Now, all of the player's totals add up make what the total team ran. Thus, the expression is;

V + 3V - 15 + V + 11 + 18 = 229

Simplifying this gives;

5V + 14 = 229

5V + 14 = 229

Subtract 14 from both sides to get;

5V = 215

Divide both sides by 5 to get;

V = 43

Thus, Vernon ran a total of 43 yards.

Total run by Reggie is:

R = 3(43) -15

R = 129 - 15

R = 114 yards.

Total run by Larry:

L = V + 11

L = 43 + 11

L = 54 yards.

Thus, we conclude that Vernon Variable ran 43 yards, Reggie Root ran 114 yards, Larry Linear ran 54 yards, and Quentin ran 18 yards.

Read more about Linear Equation Word Problem at; brainly.com/question/21405634

#SPJ1

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By using the error in circumference relation to error in radius by:

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The maximum error in surface area is simplified as:

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dA=8πr×(dC÷(2π))

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Substitute the value of r=C÷(2π) in above and get

dA=4dC×(C÷2π)

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Here, C=76cm and dC=0.5cm.

Substitute this in above as

dA=(2×76×0.5)÷π

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Find relative error as the relative error is between the value of the Area and the maximum error, therefore:

\begin{aligned}\frac{dA}{A}&=\frac{8\pi rdr}{4\pi r^2}\\ \frac{dA}{A}&=\frac{2dr}{r}\end

As above its found that r=C÷(2π) and r=dC÷(2π).

Substitute this in the above

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Hence, the maximum error in the calculated surface area with the circumference of a sphere was measured to be 76 cm with a possible error of 0.5 cm is 24.19cm² and the relative error is 0.0132.

Learn about relative error from here brainly.com/question/13106593

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