A rope is 50 meters long. It is cut into two pieces and one piece is 8 meters longer than the other. What are the two lengths?
1 answer:
You might be interested in
Given the Perimeter of the Lot is 436 ft, and an inner building whose base measures 122ft by 60ft. The key in the problem is that the easement is uniform.
Let X be the measure of easement. See attached
You will derive the equation where if you add X on the length of the inner rectangle length and width, you have below equation
P = 2L + 2W
P = 2*(122+2X) + 2*(60+2x)
436 = 244+4x + 120+4x
Rearranging the equation by shifting all constant on one side.
8x = 436-244-120
8x = 72
x=9 ft
Answer is 9ft easement for with respect to the base of the inner building.
Let X be the measure of easement. See attached
You will derive the equation where if you add X on the length of the inner rectangle length and width, you have below equation
P = 2L + 2W
P = 2*(122+2X) + 2*(60+2x)
436 = 244+4x + 120+4x
Rearranging the equation by shifting all constant on one side.
8x = 436-244-120
8x = 72
x=9 ft
Answer is 9ft easement for with respect to the base of the inner building.
To solve this problem we can use system of equations.
First equation can be
321=x+y
where x - ppl between 4-17 and y - adult
second equation can be as follows
1590=6*y+4*x
So we have
From first equation we get value of x in times of y
x=321-y
Now we can substitute it into second eq.
1590=6y+4(321-y)
now simply if
1590=6y+1284-4y /-1284 both sides
306=2y /:2 divide both sides by 2
y=153
Now we can back substitude
x=321-153
x=168
So we get result
First equation can be
321=x+y
where x - ppl between 4-17 and y - adult
second equation can be as follows
1590=6*y+4*x
So we have
From first equation we get value of x in times of y
x=321-y
Now we can substitute it into second eq.
1590=6y+4(321-y)
now simply if
1590=6y+1284-4y /-1284 both sides
306=2y /:2 divide both sides by 2
y=153
Now we can back substitude
x=321-153
x=168
So we get result