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Nitella [24]
2 years ago
5

Seraphina is driving two hours to visit her family. For the first hour, she traveled at a speed of 60 miles per hour. Then, in t

he second hour, she traveled at a speed of 68 miles per hour. What is the percentage increase of Seraphina's speed? If necessary, round to the nearest tenth of a percent.
Mathematics
1 answer:
balu736 [363]2 years ago
6 0

Answer:

Step-by-step explanation:

percent change = (amount of change)/(original amount) × 100%

... = (73 -60)/60 × 100%

... = 13/60 × 100%

... ≈ 0.216667 × 100%

... ≈ 21.7%

Sarah's speed in the second hour was about 21.7% greater than in the first hour.

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\boxed{x = 7, y = 9, z = 68}

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I will use these three:

\begin{array}{lrcll}(1) & 8x + 13y +7 & = & 180 & \\(2)& 9x - 7 + 13y +7 & = & 180 & \\(3)& 8x + 5y - 11 + z & = & 180 &\text{We can rearrange these to get:}\\(4)& 8x + 13y & = & 173 &\\(5) & 9x + 13y & = & 180 & \\(6)& 8x + 5y + z & = & 169 & \\(7)& x & = & \mathbf{7} & \text{Subtracted (4) from (5)} \\\end{array}

\begin{array}{lrcll}& 8(7) + 13y & = & 173 & \text{Substituted (7) into (4)} \\& 56 + 13y & = & 173 & \text{Simplified} \\& 13y & = & 117 & \text{Subtracted 56 from each side} \\(8)& y & =& \mathbf{9}&\text{Divided each side by 13}\\& 8(7) + 5(9) + z & = & 169 & \text{Substituted (8) and (7) into (6)} \\& 56 + 45 + z& = & 169 & \text{Simplified} \\& 101 + z& = & 169 & \text{Simplified} \\&z& = & \mathbf{68} & \text{Subtracted 101 from each side}\\\end{array}

\boxed{\mathbf{ x = 7, y = 9, z = 68}}

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