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
We rewrite the expresion:
((t+3)/(t+4))*(1/(t^2+7t+12))
We have then
((t+3)/(t+4))*(1/((t+4)*(t+3)))
Rewriting tge expression:
(1/(t+4)^2)
Answer:
(1/(t+4)^2)
opcion 3
Answer:
15 a^2 + -10 a b
Step-by-step explanation:
Expand the following:
5 a (-2 b + 3 a)
Hint: | Distribute 5 a over -2 b + 3 a.
5 a (-2 b + 3 a) = 5 a (-2 b) + 5 a (3 a):
-2 5 a b + 5 3 a a
Hint: | Combine products of like terms.
5 a×3 a = 5 a^2×3:
5×3 a^2 - 2×5 a b
Hint: | Multiply 5 and 3 together.
5×3 = 15:
15 a^2 - 2×5 a b
Hint: | Multiply 5 and -2 together.
5 (-2) = -10:
Answer: 15 a^2 + -10 a b
Answer:
45:63
Step-by-step explanation:
7+5=12
12u-->108
1u--> 9
5u-->45
7u-->63
Answer:
x=8
base side of isosceles triangle
y=√{8²+8²}=8√2