What you have to do is figure out how many half inches you need to have to figure out how many half inches equal 5 yards. Therefore if there are 12 inches in a foot, then that would be 24 half inches. So..... three feet equal 1 yards. So multiply 3 by 24. 3x24=72 so you would need 72 half inches
Answer:
D,0,2,-2
Step-by-step explanation:
2x^5-3x^3-20x=0
x(2x^4-3x^2-20)=0
x=0
or 2x^4-3x^2-20=0
put x²=t
2t²-3t-20=0
-20×2=-40
8-5=3
8×-5=-40
2t²-(8-5)t-20=0
2t²-8t+5t-20=0
2t(t-4)+5(t-4)=0
(t-4)(2t+5)=0
t=4
x²=4
x=2,-2
t=-5/2
x²=-5/2
it gives imaginary root. so real rational roots are 0,2,-2
y = x + 1
Sammy wants to train in the gym. Registration in a city gym costs $ 1. The gym fee is $ 1 per day.
We define the variables:
x = number of days.
y = full payment of the gym after x days.
Answer:
The slope represents the cost per day of the gym.
m = 1 $ / day
The intersection with the y-axis represents the inscription in the gym.
b = 1 $
Answer:
The length and width of the plot that will maximize the area of the rectangular plot are 54 ft and 27 ft respectively.
Step-by-step explanation:
Given that,
The length of fencing of the rectangular plot is = 108 ft.
Let the longer side of the rectangular plot be x which is also the side along the river side and the width of the rectangular plot be y.
Since the fence along the river does not need.
So the total perimeter of the rectangle is =2(x+y) -x
=2x+2y-y
=x+2y
So,
x+2y =108
⇒x=108 -2y
Then the area of the rectangle plot is A = xy
A=xy
⇒A= (108-2y)y
⇒ A = 108y-2y²
A = 108y-2y²
Differentiating with respect to x
A'= 108 -4y
Again differentiating with respect to x
A''= -4
For maximum or minimum, A'=0
108 -4y=0
⇒4y=108

⇒y=27.

Since at y= 27, A''<0
So, at y=27 ft , the area of the rectangular plot maximum.
Then x= (108-2.27)
=54 ft.
The length and width of the plot that will maximize the area of the rectangular plot are 54 ft and 27 ft respectively.