Answer:
The width is 50 yards and the length is 141 yards.
Step-by-step explanation:
Let's call: L the length of the field and W the width of the field.
From the sentence, the perimeter of the rectangular playing field is 382 yards we can formulate the following equation:
2L + 2W = 382
Because the perimeter of a rectangle is the sum of two times the length with two times the width.
Then, from the sentence, the length of the field is 9 yards less than triple the width, we can formulate the following equation:
L = 3W - 9
So, replacing this last equation on the first one and solving for W, we get:
2L + 2W = 382
2(3W - 9) + 2W = 382
6W -18 +2W = 382
8W - 18 = 382
8W = 382 + 18
8W = 400
W = 400/8
W = 50
Replacing W by 50 on the following equation, we get:
L = 3W - 9
L = 3(50) - 9
L = 141
So, the width of the rectangular field is 50 yards and the length is 141 yards.
Answer:
10x=log6-3
Step-by-step explanation:
Answer:
x=10, y = -20
Step-by-step explanation:
y = –5x + 30
x = 10
Substitute the second equation into the first
y = –5*10 + 30
y = -50 +30
y = -20
x=10, y = -20
For this case we have the following inequality:
2 ≥ 4 - v
The first thing we must do in this case is to clear the value of v.
We have then:
v ≥ 4 - 2
v ≥ 2
Therefore, the solution set is given by:
[2, inf)
Answer:
See attached image.
