Answer:
<h3>25m</h3>
Step-by-step explanation:
Perimeter of the rectangular pasture P = 2(L+W)
Area A = LW
L is the length
W is the width
Given
Perimeter = 110m
Area = 750m
If the length of the pasture is 40m longer than the width, then L = W+40
110 = 2L+2w
55 = L+W .....1
750 = LW.....2
Solving simultaneously
from 1; L = 55-W
substitute into 2;
750 = (55-W)W
750 = 55W-W²
-W²+55W -750 = 0
W²-55W+750 = 0
(W²-25W)-(30W+750) = 0
W(W-25)-30(W-25) = 0
(W-25)(W-30) = 0
W-25 = 0 and W-30 = 0
w = 25m and 30m
Since L = 55-W
L = 55-25 = 30m and;
L = 55-30 = 25m
Since we are told that length id longer than the width then, the width we are going to use is 25m
A = (14)(18) + 1/2(18)(13)
A = 252 + 117
A = 369
answer
369 m^2
Four and 23 hundredths
4.23
I hope this helps! :D (Sorry if these are wrong.)
Answer:
see the explanation
Step-by-step explanation:
Let
x ------> the next grade
we know that
To find the average grade sum the grades by the number of math test
so

solve for x
Combine liker terms in the right side

Multiply by 5 both sides

Subtract 372 both sides

-----> this is not make sense
The grade cannot be a negative number
so
For x=0
The minimum average grade is

The answer is C.
To solve these types of equations you first need to make sure that you equation is equal to 0. In this case all of the numbers are already on the same side so no work needs to be done there. Then we can get a, b and c values for the quadratic equation by looking at the coefficients.
a = 4 (number attached to x^2)
b = -6 (number attached to x)
c = 1 (number with no variable attached)
Now we can put this into the quadratic equation.

