Question asked:Jacob and Sarah are saving money to go on a trip. They need at least $1975 in order to go. Jacob mows lawns and Sarah walks dogs to raise money. Jacob charges $25 each time he mows a lawn and Sarah charges $15 each time she walks a dog. The number of dog walks that Sarah has scheduled is no more than four times the number of lawns Jacob has scheduled to mow. Sarah will walk at least 50 dogs.Write a set of constraints to model the problem, with x representing the number of lawns mowed and y representing the number of dogs walked.
My answer:
They need at least $1975. Jacob charges $25 each time he mows a lawn and Sarah charges $15 each time she walks a dog.
25x + 15y ≥ 1975
The number of dog walks that Sarah has scheduled is no more than four times the number of lawns Jacob has scheduled to mow.
y ≤ 4x
Sarah will walk at least 50 dogs.
y ≥ 50
MN is the mid-line of a trapezoid HJLK.<span>
The length of the mid-line of a trapezoid is half of the sum of the lengths of its bases </span>⇒
<span>
(KL + HJ)/2 = MN
(KL + 45)/2 = 28
KL + 45 = 28 * 2
KL + 45 = 56
KL = 56 - 45
KL = 11
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Answer:
I think they are similar by SAS.
Step-by-step explanation:
Both triangles may not be the same size, but they have similar sides and similar angles.
U didn't mess up....
V = pi * r^2 * h....divide both sides by (pi)r^2
V / (pi)r^2 = h
It equals 276 but about 300
