EQUATION:
(30+2x)(40+2x) = 1800 sq ft.
4(15+x)(20+x) = 1800
x^2+35x+300=450
x^2+35x-150=0
Use quadratic formula to get:
x=[-35+sqrt(35^2-4*-150)]/2
x=[-35+-sqrt1825]/2
x=[-35+-42.72]/2
x=7.72/2
x=3.86 ft
Can someone set this up in equations so i can solve them i would greatly appreciate it. THanks.
Problem #3
The base of a ladder is 14 feet away from the wall. The top of the ladder is 17 feet from the floor. Find the length of the ladder to the nearest thousandth.
Problem #4
A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What must be the width of the walkway to the nearest thousandth?
112 because im smart im 10 and im in the 11th grade
oh sorry the answer is 2,000
the next one is 4,000
-x+y=6
I do not want the answers just the equations.
3) Think about what shape that would make. If you're not sure, draw it out. The formula? a^2 + b^2 = .... ?
4) Again, this may help to draw out a picture of it. It would look like a rectangle inside of a rectangle. Now that you see that much, what's the area of the garden? We know it's 30ft x 40ft - which is 1200 ft^2
The area of the whole place is 1800 ft^2
So subtract 1800 - 1200 and we see the area of the walkway is 600 ft^2
Try to figure it out from there and see what you come up with.
Matt
2 answers
Problem 4 solution:
EQUATION:
(30+2x)(40+2x) = 1800 sq ft.
4(15+x)(20+x) = 1800
x^2+35x+300=450
x^2+35x-150=0
Use quadratic formula to get:
x=[-35+sqrt(35^2-4*-150)]/2
x=[-35+-sqrt1825]/2
x=[-35+-42.72]/2
x=7.72/2
x=3.86 ft
EQUATION:
(30+2x)(40+2x) = 1800 sq ft.
4(15+x)(20+x) = 1800
x^2+35x+300=450
x^2+35x-150=0
Use quadratic formula to get:
x=[-35+sqrt(35^2-4*-150)]/2
x=[-35+-sqrt1825]/2
x=[-35+-42.72]/2
x=7.72/2
x=3.86 ft