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serious [3.7K]
3 years ago
9

Please solve this problem for me.

Mathematics
2 answers:
xenn [34]3 years ago
7 0

He might have done it backwards.

1/63

and could be 63 miles in 1 hour

63/1.

Because the graph looks good. (i think)

iragen [17]3 years ago
6 0

Answer:

It is 63 or 63/1. Billy used change in x over change in y, and the correct formula for slope is the change in y over the change in x.

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Answer18:

The quadrilateral ABCD is not a parallelogram

Answer19:

The quadrilateral ABCD is a parallelogram

Step-by-step explanation:

For question 18:

Given that vertices of a quadrilateral are A(-4,-1), B(-4,6), C(2,6) and D(2,-4)

The slope of a line is given m=\frac{Y2-Y1}{X2-X1}

Now,

The slope of a line AB:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{6-(-1)}{(-4)-(-4)}

m=\frac{7}{0}

The slope is 90 degree

The slope of a line BC:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{6-6}{(-4)-(-1)}

m=\frac{0}{(-3)}

The slope is zero degree

The slope of a line CD:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{(-4)-6}{2-2}

m=\frac{-10}{0}

The slope is 90 degree

The slope of a line DA:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{(-1)-(-4)}{(-4)-(2)}

m=\frac{3}{-6}

m=\frac{-1}{2}

The slope of the only line AB and CD are the same.

Thus, The quadrilateral ABCD is not a parallelogram

For question 19:

Given that vertices of a quadrilateral are A(-2,3), B(3,2), C(2,-1) and D(-3,0)

The slope of a line is given m=\frac{Y2-Y1}{X2-X1}

Now,

The slope of a line AB:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{2-3}{3-(-2)}

m=\frac{-1}{5}

The slope of a line BC:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{(-1)-2}{2-3}

m=\frac{-3}{-1}

m=3

The slope of a line CD:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{0-(-1)}{(-3)-2}

m=\frac{-1}{5}

The slope of a line DA:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{3-0}{(-2)-(-3)}

m=3

The slope of the line AB and CD are the same

The slope of the line BC and DA are the same

Thus, The quadrilateral ABCD is a parallelogram

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Answer:

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