We can solve this by looking at pairs that equal 60.
6 and 10 get to 60 when multiplied. Let's see if it fits in as a perimeter.
12+20=32
So the width and length of the room is 6 meters and 10 meters.
Answer:
If the perimeter of the rectangle is 60 feet, how long are the base and height of Herman's rectangle? Show how you know.
Step-by-step explanation:
B=4x|(4.10) x+x+4x+4*=60.
Answer:
Step-by-step explanation:
Given
Area of a floor = 2040ft²
If Its length is 36ft. longer than its width, then L = W+36
Area = LW
L is the length
W is the width
A < W(W+36)
2040 < W²+36W
0<W²+36W-2040
W²+36W-2040 >0
W = -36±√36²-4(-2040)/2
W = -36±√1296+8160)/2
W = -36±√9456/2
W = -36±97.25/2
W = -36+97.25/2
W = 61.24/2
W > 30.62ft
Since L = W+36
L > 30.62+36
L > 66.62ft
The width of the floor is 30.62ft
The length is 66.62ft
The mathematical sentence would represent the given situation is the width is greater than 30.62ft while the length must be greater than 66.62ft
The possible dimension of the floor is 31ft by 67 ft
The possible areas is 31*67 = 2077ft²
It won't be realistic to get an area of 144sqft because the initial area is greater than 144. Hence the feasible area will be greater than 2040ft²
Answer:
It would be 4(X + 3)
Step-by-step explanation:
There are 4 Xs and 12 ones, so the equation is 4X + 12 or 4(X + 3).
Answer: Link the graphs and more information, cannot answer the question because of insufficient information.
Step-by-step explanation:
