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3 years ago

PLEASE HELP! HELP! Plot the points (−3, 0) and (3, 5) on the coordinate plane.

Mathematics

1 answer:

iren [92.7K]3 years ago

6 0

(Image below)

Keep in mind that the first number in a coordinate is always on the x-axis (the horizontal line) and the second number is on the y-axis (the vertical line).

If the question asks us to plot (-3, 0), we need to find -3 on the horizontal line and find 0 on the vertical line.

Since 0 means that we go 0 units up, we can just plot the point at -3 on the x-axis.

Next, we need to plot (3, 5). First, let's find the first coordinate by locating 3 on the horizontal axis. Now, we need to find 5 up from that point, because our y-coordinate is 5.

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Given Information:

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Required Information:

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Answer:

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Step-by-step explanation:

Recall that the rate is given by

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$\frac{160}{(r-4)} = \frac{160}{r} + 2 \\\\$

Simplify the equation

$\frac{160}{(r-4)} - \frac{160}{r} = 2 \\\\\frac{160r - 160(r-4)}{r(r-4)} = 2 \\\\\frac{160r - 160r+640)}{r(r-4)} = 2 \\\\160r - 160r+640 = r(r-4) (2) \\\\640 = r(r-4) (2) \\\\640 = (r^2 - 4r) (2) \\\\640 = 2r^2 - 8r \\\\2r^2 - 8r-640 =0 \\\\2(r^2 - 4r-320) =0 \\\\r^2 - 4r-320 =0 \\\\$

Now we are left with a quadratic equation.

We may solve the quadratic equation using the factorization method

$r^2 - 4r-320 =0 \\\\r^2-20r+16r-320=0 \\\\r(r-20)+16(r-20)=0 \\\\(r-20) (r+16)=0 \\\\$

So,

$(r-20) = 0 \\\\r = 20 \\\\$

$(r+16)=0 \\\\r = -16 \\\\$

Since rate cannot be negative, discard the negative value of r

Therefore, the rate is

$r = 20 \: mph$

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