Answer:
The largest total area that can be enclosed will be a square of length 272 yards.
Step-by-step explanation:
First we get the perimeter of the large rectangular enclosure.
Perimeter of a rectangle =2(l + w)
Perimeter of the large rectangular enclosure= 1088 yard
Therefore:
2(L+W)=1088
The region inside the fence is the area
Area: A = LW
We need to solve the perimeter formula for either the length or width.
2L+ 2W= 1088 yd
2W= 1088– 2L
W = 
W = 544–L
Now substitute W = 544–L into the area formula
A = LW
A = L(544 – L)
A = 544L–L²
Since A is a quadratic expression, we re-write the expression with the exponents in descending order.
A = –L²+544L
Next, we look for the value of the x coordinate
L=272 yards
Plugging L=272 yards into the calculation for area:
A = –L²+544L
A(272)=-272²+544(272)
=73984 square yards
Thus the largest area that could be encompassed would be a square where each side has a length of 272 yards and a width of:
W = 544 – L
= 544 – 272
= 272 yards
Answer:
A, 4, -1
It's reflected over the y-axis, so it goes to the right side.
Answer:c
Step-by-step explanation:
3x - 3y + 9 = 0
The y-intercept is the point on the graph where it crosses the y-axis, and has coordinates of (0, b). It is also the value of y when x = 0.
To solve for the y-intercept, set x = 0:
3(0) - 3y + 9 = 0
3(0) - 3y + 9 = 0
Subtract 9 from both sides:
- 3y + 9 - 9 = 0 - 9
- 3y = -9
Divide both sides by -3 to solve for y:
-3y/-3 = -9/-3
y = 3
Therefore, the y-intercept is (0, 3).
The x-intercept is the point on the graph where it crosses the x-axis, and has coordinates of (a, 0). It is also the value of x when y = 0.
To solve for the x-intercept, set y = 0:
3x - 3(0)+ 9 = 0
3x -0 + 9 = 0
Subtract 9 from both sides:
3x + 9 - 9 = 0 - 9
3x = -9
Divide both sides by 3 to solve for x:
3x/3 = -9/3
x = -3
Therefore, the x-intercept is (-3,0).
The correct answers are:
Y-intercept = (0, 3)
X-intercept = (-3, 0)
9514 1404 393
Answer:
- 3n+6 (n = smallest)
- 3n (n = middle)
Step-by-step explanation:
The usual method for doing this is to let n represent the smallest one. Then the three integers are ...
n, n+2, and n+4
and their sum is ...
(n) +(n+2) +(n+4) = 3n+6 . . . . sum of 3 consecutive odd integers (n = smallest)
_____
Personally, for consecutive number problems, I prefer to let the variable represent the average value. If n is the average value of 3 consecutive odd integers, is is the middle integer. Of course, the sum will be 3 times the average:
(n-2) +(n) +(n+2) = 3n . . . . sum of 3 consecutive odd integers (n = middle one)
