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kiruha [24]
3 years ago
15

Two cars leave towns 400 kilometers apart at the same time and travel toward each other. One car's rate is 14 kilometers per hou

r less than the other's. If they
meet in 2 hours, what is the rate of the slower car?
Do not do any rounding.
Mathematics
1 answer:
Scorpion4ik [409]3 years ago
3 0

Answer:

Step-by-step explanation:

v*2+(v-14)*2=400

2v+2v-28=400

4v=400+28

4v=428

v=107 km/h

speed of slowest car=107-14=93 km/h

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Answer:

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Step-by-step explanation:

Given

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Required

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First, we need to determine the inequalities of the system.

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x\ +\ y\ge 20

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Divide through by 0.01

\frac{0.01y}{0.01} \le \frac{0.80}{0.01} -\frac{ 0.05x}{0.01}

y \le \frac{0.80}{0.01} -\frac{ 0.05x}{0.01}

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The resulting inequalities are:

y \ge 20 - x

y \le 80 -5x

The two inequalities are plotted on the graph as shown in the attachment.

y \ge 20 - x --- Blue

y \ge 80 -5x --- Green

Point A on the attachment are possible solutions

At A:

(x,y) = (15,5)

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