Answer:Area of part 1:
A1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2
Area of part 2:
A2=length*width=(8+2)*1=10 ft2
Area of part 3:
A3=length*width=8*2=16 ft2
Total shaded area : A1+A2+A3=41 ft2
Total area of the reactangle : A=16*8=128 ft2
Total area of the nonshaded region : 128-41=87 ft2
Step-by-step explanation: Plz make brainliest
Answer:
Angles G and H are congruent.
Step-by-step explanation:
The length of a segment is the distance between its endpoints.

- AB and CD are not congruent
- AB does not bisect CD
- CD does not bisect AB
<u>(a) Length of AB</u>
We have:


The length of AB is calculated using the following distance formula

So, we have:


Simplify

<u>(b) Are AB and CD congruent</u>
First, we calculate the length of CD using:

Where:


So, we have:



By comparison

Hence, AB and CD are not congruent
<u>(c) AB bisects CD or not?</u>
If AB bisects CD, then:

The above equation is not true, because:

Hence, AB does not bisect CD
<u>(d) CD bisects AB or not?</u>
If CD bisects AB, then:

The above equation is not true, because:

Hence, CD does not bisect AB
Read more about lengths and bisections at:
brainly.com/question/20837270
Answer: <u>110 feet per second</u>
Step-by-step explanation:
In this problem we will assume that the car at 45 miles per hour is moving into the x direction with a high of 100 feet, and the train is going in the y direction, so they trajectories will made an angle of 90º.
Now we can calculate the speed of the trains in feet per second so:


So we can make a right triangle with sides 66 and 88 and the hypotenuse will be the rate that the trains will separate per second so:


Is important to have in mind that the initial high is not going to change how fast the trains will separate, however, if we are going to calculate the distance we should have it in the calculations.