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Otrada [13]
4 years ago
5

PLEASE HELP 25 POINTS!!! WILL GIVE BRAINLIEST NEED HELP ASAP

Mathematics
1 answer:
leva [86]4 years ago
3 0

Answer:

Bus B travels faster.

Step-by-step explanation:

Bus A travels according to the equation

y = 125/2 x

This equation is in the form y = mx + b, where m = 125/2, and b = 0.

m is the slope, so the slope is 125/2 which equals 62.5

The slope is the rate of change. Since y is distance in miles, and x is time in hours, the slope is the speed in miles per hour.

The speed of Bus A is 62.5 miles per hour.

Bus B travels according to the graph. We can find the slope by using easy-to-read points on the graph. The slope is the speed in miles per hour.

Look for two points on grid line intersections since they are the easiest ones to read. Let's use point (0, 0) and point (3, 200). Now that we have 2 points, we can find the slope.

slope = \dfrac{y_2 - y_1}{x_2 - x_1}

slope = \dfrac{200 - 0}{3 - 0}

slope = \dfrac{200}{3} = 66.67

The speed of Bus 2 is 66.67 miles per hour.

Answer: Bus B travels faster.

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<u>The three steps are</u>

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<u>Correct solution should be:</u>

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<u>This would give next steps:</u>

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<u>Answer options:</u>

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Step 3 wasn't performed yet, there was step 1 and 2.

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since the slower pump takes L hours, the faster pump takes then L - 5 hours.

we know both pumps together take 3 hours to do <u>half of the pool</u>, so that means that to fill up <u>the whole pool it takes them 6 hours</u>.

since the slower pump can do it alone in L hours, in 1 hour it has done 1/L of the whole thing.

likewise, since the faster pump can do it in L-5 hours, in 1 hour alone it has done 1/(L-5) of the whole job.


\bf \stackrel{\textit{slower pump's rate}}{\cfrac{1}{L}}~~~~+\stackrel{\textit{faster pump's rate}}{\cfrac{1}{L-5}}~~~~=~~~~\stackrel{\textit{total done in 1 hour}}{\cfrac{1}{6}} \\\\\\ \textit{let's multiply both sides by }\stackrel{LCD}{6(L)(L-5)}\textit{ to do away with the denominators} \\\\\\ 6(L)(L-5)\left( \cfrac{1}{L}+\cfrac{1}{L-5} \right)=6(L)(L-5)\left( \cfrac{1}{6} \right)


\bf 6(L-5)+6L=(L)(L-5)\implies 6L-30+6L \\\\\\ 12L-30=L^2-5L\implies 0=L^2-17L+30 \\\\\\ 0=(L-15)(L-2)\implies L= \begin{cases} \boxed{15}\\ 2 \end{cases}


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now if L = 15, then L - 5 = 10.

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