A. 1.75+0.25x<15;x,53miles
sorry, i could not put in the or equal to signs
Answer:
10 and 15
Step-by-step explanation:
Let 'x' and 'y' are the numbers we need to find.
x + y = 25 (two numbers whose sum is 25)
(1/x) + (1/y) = 1/6 (the sum of whose reciprocals is 1/6)
The solutions of the this system of equations are the numbers we need to find.
x = 25 - y
1/(25 - y) + 1/y = 1/6 multiply both sides by 6(25-y)y
6y + 6(25-y) = (25-y)y
6y + 150 - 6y = 25y - (y^2)
y^2 - 25y + 150 = 0 quadratic equation has 2 solutions
y1 = 15
y2 = 10
Thus we have
:
First solution: for y = 15, x = 25 - 15 = 10
Second solution: for y = 10, x = 25 - 10 = 15
The first and the second solution are in fact the same one solution we are looking for: the two numbers are 10 and 15 (since the combination 10 and 15 is the same as 15 and 10).
Answer:
the answer is most likely A
Step-by-step explanation:
Area is 6000 yd2
LW = 6000
field must be 40 yards longer than its width
L = W + 40
Replace L with W+40 in 1st equation to solve for Width
(W+40)(W) = 6000
W2 + 40W - 6000 = 0
This is a quadratic but is factorable
Factors of -6000 that add to 40 are (100)(-60)
(W+100)(W-60) = 0
W = -100 or W = 60
Since the width will not be negative discard -100
The width is 60 yards
Length is W+40 = 100 yards
Answer:b
Step-by-step explanation:
