Answer: 45
Step-by-step explanation:
60 x .75. = 45
X and w is 105 degrees and y and z are 86 degrees. since line L is a straight line it would have to add up to 180 and same with line M

hope this helps! ask if you have anymore questions
(a) Length of the height is 2.732 m
(b) Length of the base is 5.466 m
<u>Explanation:</u>
An image is attached for reference.
(a)
In ΔAOB,

In ΔBGD,

According to the figure, BG = OE = 1.732 m
Height of the tent, AE = AO + OE
= 1 m + 1.732 m
= 2.732 m
(b)
DF = ?
In ΔAOB,

According to the figure, OB = GE = 1.733 m
In ΔBGD,

According to the figure, DE = DG + GE
DE = 1 m + 1.733 m
DE = 2.733 m
Length of the base, DF = 2 X DE
DF = 2 X 2.733 m
DF = 5.466 m
Answer:
The question does not have value of height (in the image)
the barn in shape of a cuboid with a half cylinder.
volume of barn = volume of cuboid + volume of cylinder/2

here,

b=6 m
l=10 m
so, 2r=6. thus, r=3,

which approximately,
141.3+60h
for storing all of the hay
141.3+60h≥600
h≥7.644
so..
check if h is less than 7.644, he would not be able to store all of the barn.
mark brainliest please.