First you plot in the y-intercept of the equation. To find the y-intercept, substitute 0 into x. -3m will cancel our giving you y=5. x=0, y=5, the first ordered pair is (0,5). Now after you plot in the y-intercept, use your slope, which is -3, to graph the points of the equation. Starting from (0,5), move down 3 spaces on the y-axis (because it’s -3) and you’ll end up at (0,2). Next move over 1 ( all slopes with just a whole number moves on the x-axis 1 since the whole number divided by 1 doesn’t change the slope number) to the right because it’s a negative linear equation so it’ll go downward. After moving right, you’ll get (1,2). Do a couple more points starting from (1,2) then the 3rd point ABD and so on to get 3 or more points to be able to draw a linear line.
Step-by-step explanation:
let us give all the quantities in the problem variable names.
x= amount in utility stock
y = amount in electronics stock
c = amount in bond
“The total amount of $200,000 need not be fully invested at any one time.”
becomes
x + y + c ≤ 200, 000,
Also
“The amount invested in the stocks cannot be more than half the total amount invested”
a + b ≤1/2 (total amount invested),
=1/2(x + y + c).
(x+y-c)/2≤0
“The amount invested in the utility stock cannot exceed $40,000”
a ≤ 40, 000
“The amount invested in the bond must be at least $70,000”
c ≥ 70, 000
Putting this all together, our linear optimization problem is:
Maximize z = 1.09x + 1.04y + 1.05c
subject to
x+ y+ c ≤ 200, 000
x/2 +y/2 -c/2 ≤ 0
≤ 40, 000,
c ≥ 70, 000
a ≥ 0, b ≥ 0, c ≥ 0.
Answer:
x/y=1/4
Step-by-step explanation:
3x+2y=3y-2
3x=3y-2y-2
3x=y-2
y=3x+2
x+y=10
x+3x+2=10
4x+2=10
4x=10-2
4x=8
x=8/4
x=2
2+y=10
y=10-2
y=8
x/y=2/8=1/4
B. 7
49 divided by 7 is 7.
