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
Answer:
The correct options are:
- g(x) is shifted three units higher than f(x).
- g(x) has a period that is half the period of f(x).
Step-by-step explanation:
We have to compare the graphs of the function:

and 
We have to select the correct options among the following:
As we know that the period of sine function is 2π.
i.e. Period of function f(x) is: 2π.
The period of sin(2 x) is π.
Hence, the period of the function g(x) function is π.
- Hence, the period of g(x) is half the period of f(x).
- Also we could observe that g(x) is shifted 3 units upward.
Answer: 600 m/s^2
Step-by-step explanation:
Use the first kinematic equation

Answer:
1/9
Step-by-step explanation:
6/9 is 1/9th bigger than 5/9
Answer:
thats a bit confusing ngl
Step-by-step explanation: