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vlada-n [284]
3 years ago
15

The distance from Charles’s house to his school is 3 miles. His father gave him a ride for 1,760 yards, and he walked the rest o

f the way. The distance that Charles walked is Either
1
1.5
or 2 miles,
or 1,760
2,000,
3,520 yards.
(Hint: 1 mile = 1,760 yards.)
Mathematics
2 answers:
slamgirl [31]3 years ago
8 0
The answer is 2 miles or 3520 yards. (they are the same answer.)

If this helps, can you give me the brainliest answer? I need it...

Thank you very much.
sergiy2304 [10]3 years ago
4 0

Answer:

2 miles or 3,520 yards.

Step-by-step explanation:

The distance from Charles's house to his school = 3 miles

1 mile = 1760 yards.

His father gave him a ride for 1,760 yards = 1 mile

He walked the rest of the way that is = 3 miles - 1 mile = 2 miles

Since,  1 mile = 1760

Therefore 2 miles = 1760 × 2 = 3,520 yards.

So Charles walked 2 miles or you can say 3,520 yards.

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Here i how I would do it:<span>f(x)=−<span>x2</span>+8x+15</span>
set f(x) = 0 to find the points at which the graph crosses the x-axis. So<span>−<span>x2</span>+8x+15=0</span>
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To find the coordinates of the maximum (it is maximum) of the graph, you take a look at the completed square method above. Since we multiplied through by -1, we need to multiply through by it again to get:<span>f(x)=31−(x−4<span>)2</span></span><span>
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How do you do this problem? I need to know how you found the answer.
Alexxandr [17]

to get the equation of a line, we simply need two points, say for the Red one ... notice in the graph the lines passes through (0,2) and (-1,6), so let's use those


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{-1-0}\implies \cfrac{4}{-1}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=-4(x-0) \\\\\\ y-2=-4x\implies \blacktriangleright y=-4x+2 \blacktriangleleft


now, for the Blue one, say let's use hmmm it passes through (0,2) and (1.6)


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{1-0}\implies \cfrac{4}{1}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=4(x-0) \\\\\\ y-2=4x\implies \blacktriangleright y=4x+2 \blacktriangleleft

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