Answer:
y = -¼│x − 5│+ 3
Step-by-step explanation:
y = a│x − h│+ k
(h, k) is the vertex of the absolute value graph. In this case, it's (5, 3).
y = a│x − 5│+ 3
One point on the graph is (1, 2). Plug in to find the value of a.
2 = a│1 − 5│+ 3
2 = 4a + 3
a = -¼
Therefore, the graph is:
y = -¼│x − 5│+ 3
Answer:
D
Step-by-step explanation:
4 + 12 + 18 + 68= 102/2 51
40 + 28 + 52 + 76= 196/2 98
51:98
The point-slope form:
![y-y_1=m(x-x_1)](https://tex.z-dn.net/?f=y-y_1%3Dm%28x-x_1%29)
m - slope
x₁, y₁ - the coordinates of a point
It passes through the points (1,-2) and (2,2).
![(1,-2) \\ x_1=1 \\ y_1=-2 \\ \\ (2,2) \\ x_2=2 \\ y_2=2 \\ \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{2-(-2)}{2-1}=\frac{2+2}{1}=4](https://tex.z-dn.net/?f=%281%2C-2%29%20%5C%5C%0Ax_1%3D1%20%5C%5C%20y_1%3D-2%20%5C%5C%20%5C%5C%0A%282%2C2%29%20%5C%5C%0Ax_2%3D2%20%5C%5C%20y_2%3D2%20%5C%5C%20%5C%5C%0Am%3D%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D%3D%5Cfrac%7B2-%28-2%29%7D%7B2-1%7D%3D%5Cfrac%7B2%2B2%7D%7B1%7D%3D4)
Answer:
The common ratio is 1.2.
After 12 weeks the plant will be 4.46 meters long.
After one year the plant will be 6,552.32 meters long.
Step-by-step explanation:
Given that a marrow plants 0.5 m long and every week it grows by 20%, to find the value of the common ratio, calculate how long the plant is after 12 weeks and comment on the predicted length of the marrow after one year must be performed the following calculations:
0.50 x 1.2 ^ X = Y
Therefore, the common ratio is 1.2.
0.50 x 1.2 ^ 12 = Y
0.50 x 8.916 = Y
4.4580 = Y
Therefore, after 12 weeks the plant will be 4.46 meters long.
0.50 x 1.2 ^ 52 = Y
0.50 x 13,104.63 = Y
6,552.32 = Y
Therefore, after one year the plant will be 6,552.32 meters long.
Answer:
x = 125 ft and y = 250/3 ft
Step-by-step explanation:
Let assume that,
x be the length of the northern part of the fence (parallel to the north wall)
y be the length of the western and eastern pieces of the fence
As well as farmer has $4000 to spend hence we can write,
8x + 8y + 4y = 4000
8x + 12y = 4000
Hence we can say that,
from the above equation we can write
![y=\frac{4000}{12} -\frac{2x}{3}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B4000%7D%7B12%7D%20-%5Cfrac%7B2x%7D%7B3%7D)
We know that area
A=xy.
![A(x) =\frac{500x}{3}-\frac{2x^2}{3}](https://tex.z-dn.net/?f=A%28x%29%20%3D%5Cfrac%7B500x%7D%7B3%7D-%5Cfrac%7B2x%5E2%7D%7B3%7D)
we can write
![A'(x) = \frac{500}{3} -\frac{4x}{3}](https://tex.z-dn.net/?f=A%27%28x%29%20%3D%20%5Cfrac%7B500%7D%7B3%7D%20-%5Cfrac%7B4x%7D%7B3%7D)
equating it 0 we get
![A'(x) = \frac{500}{3} -\frac{4x}{3}=0](https://tex.z-dn.net/?f=A%27%28x%29%20%3D%20%5Cfrac%7B500%7D%7B3%7D%20-%5Cfrac%7B4x%7D%7B3%7D%3D0)
![x=125](https://tex.z-dn.net/?f=x%3D125)
Also,
![A"(x) = -\frac{4}{3}](https://tex.z-dn.net/?f=A%22%28x%29%20%3D%20-%5Cfrac%7B4%7D%7B3%7D)
which is less than zero.
we can see that A''(x) is always less than 0 hence using second derivative test we can say that x = 125 is a maximum point.
now, solving for y we get ![y= \frac{250}{3}](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B250%7D%7B3%7D)
Hence we can say that dimensions for the plot that would enclose the most area is,
x = 125 ft and y = 250/3 ft