Answer: 81 - 74 = 7
Have a great day!
Step-by-step explanation:
Equation of straight line is y=mx+c
choose any two points on straight line
for me I choose:(-3,11) and (3,-1)
use these two points to find gradient,m.
m= (-1-11)/(3-(-3))
m= -2
now, y=-2x+c
choose any point on the straight line
I choose point (3,-1)
sub the point into the equation to find c
-1=-2(3)+c
c=5
equation: y=-2x+5
Add the equations,just the way they appear there.
-- Add the top 'x' to the bottom 'x'. Write the sum under the 'x's.
-- Add the '+y' and the '-y'. Write the sum under the 'y's.
-- Add the '2' and the '4'. Write the sum under them, with n " = " sign
before it.
You should now have an equation with only 'x' in it and no 'y'.
You can easily solve that one and find out the value of 'x'.
Once you know the value of 'x', go back to either one of the original
equations, and plug the number-value of 'x' in place of 'x'.
You'll then have an equation with only 'y' in it and no 'x'.
You can easily solve that one and find out the value of 'y'.
Problem 1
<h3>Answer:
6.7</h3>
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Work Shown:
The two points are
and 
Apply the distance formula to get the following

The distance between the two endpoints is roughly 6.7 units. This is the same as saying the segment is roughly 6.7 units long.
======================================================
Problem 2
<h3>Answer: 3.6</h3>
----------------
Work Shown:
We'll use the distance formula here as well.
This time we have the two points
and 
The distance between them is...

This distance is approximate.
Let's call the length of the rectangular fence
and the width of the rectangular fence
.
Based on the information in the problem, we can make two equations, based on the formulas for perimeter and area:


Now, let's substitute in the values we are given in the problem:


Now, let's solve the system using substitution.

- Solve for
from the first equation

- Substitute this value into the second equation


- Multiply both sides of the equation by


- Subtract
from both sides of the equation

and 
- Use the Zero Product Property and solve both factors


We are given two possible lengths. However, the funny thing is that both produce the same dimensions: 24 ft by 10 ft:


Thus, our answer is Choice C, or 10 feet by 24 feet.