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shtirl [24]
3 years ago
6

a football player runs across the 30-yard line at an angle. He continues in a straight line and crosses the goal line at the sam

e angle. Describe two parallel lines amd a transversal in the diagram.

Mathematics
2 answers:
Vlad [161]3 years ago
8 0

Answer:

The 30 yard line and the goal line are the parallel lines and the path of the runner is the transversal

Step-by-step explanation:

Consider the provided information.

Transverse: Transverse is a line that cuts across two or more lines.

Parallel line: Parallel lines are lines which lying at equidistant in the same plane and do not meet.

It is given that  football player runs across the 30-yard line at an angle and crosses the goal line at the same angle.

From the figure is clear that 30 yard line is parallel to the goal line where the path of the runner is transversal.

For better understanding refer the figure 1.

Hence, the 30 yard line and the goal line are the parallel lines and the path of the runner is the transversal

vovangra [49]3 years ago
4 0

To solve this problem you must know the following information:

1. Two lines are parallel when they are equidistant and never intersect each other, therefore, the have equal slope.

2. A transversal line is a line that cut two or more lines at different points.

3. Based on the information shown above, you have that the 30-yard line and  the goal line are equidistant, they do not intersect each other and both have the same slope. Therefore, they are parallel.

4. The straight line with an angle intersect the parallel lines, therefore, it is the transversal.

The answer is:  The 30-yard line and  the goal line are parallel. The straight line that shows the path of the football player is the transversal line.

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A rectangular piece of metal is 15 in longer than it is wide. Squares with sides 3 in long are cut from the four corners and the
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Answer:

The original dimensions of the piece of metal are

Length: 36 inches

Width: 21 inches

Step-by-step explanation:

Let

x ----> the original length pf the piece of metal

y ----> the original width pf the piece of metal

we know that

x=y+15 ----> equation A

The volume of the box is equal to

V=LWH

we have

V=1,350\ in^3

L=(x-3-3)=(x-6)\ in\\W=(y-3-3)=(y-6)\ in

H=3\ in

substitute in the formula of volume

1,350=(x-6)(y-6)(3) ----> equation B

substitute equation A in equation B

1,350=(y+15-6)(y-6)(3)

solve for y

450=(y+9)(y-6)

y^2-6y+9y-54=450\\y^2+3y-504=0

Solve the quadratic equation by graphing

using a graphing tool

The solution is y=21

Find the value of x

x=21+15=36

therefore

The original dimensions of the piece of metal are

Length: 36 inches

Width: 21 inches

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