1. BC is 1+3=4 (units)
2. AD is 3+5= 8 (units)
3. Draw the altitude CH from point C, as shown in the figure.
4. Then triangle CHD is a right triangle, with hypothenuse CD, and sides HD and CH.
5.
6. AB=CD=4.47 because the trapezoid is isosceles.
7. P= BC+AD+AB+CD=4+8+4.47+4.47=20.9 (units)
Problem 1
<h3>Answer: C. lines a and b</h3>
Explanation: Circle or highlight the angles 10 and 15. They are alternate interior angles with line d being the transversal cut. It might help to try to erase line c to picture the transversal line d better. With d as the transversal, and angles 10 and 15 congruent, this must mean lines a and b are parallel by the alternate interior angle theorem converse.
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Problem 2
<h3>Answer: D. no</h3>
Explanation: The angles at the top are 32 degrees, 90 degrees, and x degrees which is the missing unmarked angle at the top (all three angles are below line m). The three angles must add to 180 to form a straight angle
32+90+x = 180
x+122 = 180
x = 180-122
x = 58
The missing angle is 58 degrees. This is very close to 57 degrees at the bottom. Though we do not have an exact match. This means lines m and n are not parallel. The alternate interior angles must be congruent for m and n to be parallel, as stated earlier in problem 1.
The answer is 4y^4.
Comment if you want me to explain.
Answer:
Step-by-step explanation:
B. The quotient of - 36 and - 6
E. - 36 distributed into groups of - 6
F. - 36 divided by - 6
For number 14 the answer is 10:45
