Answer:
x = -5, and y = -6
Step-by-step explanation:
Suppose that we have two equations:
A = B
and
C = D
combining the equations means that we will do:
First we multiply both whole equations by constants:
k*(A = B) ---> k*A = k*B
j*(C = D) ----> j*C = j*D
And then we "add" them:
k*A + j*C = k*B + j*D
Now we have the equations:
-x - y = 11
4*x - 5*y = 10
We want to add them in a given form that one of the variables cancels, so we can solve it for the other variable.
Then we can take the first equation:
-x - y = 11
and multiply both sides by 4.
4*(-x - y = 11)
Then we get:
4*(-x - y) = 4*11
-4*x - 4*y = 44
Now we have the two equations:
-4*x - 4*y = 44
4*x - 5*y = 10
(here we can think that we multiplied the second equation by 1, then we have k = 4, and j = 1)
If we add them, we get:
(-4*x - 4*y) + (4*x - 5*y) = 10 + 44
-4*x - 4*y + 4*x - 5*y = 54
-9*y = 54
So we combined the equations and now ended with an equation that is really easy to solve for y.
y = 54/-9 = -6
Now that we know the value of y, we can simply replace it in one of the two equations to get the value of x.
-x - y = 11
-x - (-6) = 11
-x + 6 = 11
-x = 11 -6 = 5
-x = 5
x = -5
Then:
x = -5, and y = -6
23.75 miles, because you would take 4 and 3/4 (4.75) and multiply by 5 :)
The right system of equations to describe the situation would
be on the form:
x1 = 8000 + y1*t
and
x2 = 8000 + y2*t
where x1 and x2 represents the total money of Imogene and her
friend respectively at the end of t years.
Now for the value of amount earned, y1 and y2:
y1=8000*0.08
y2=2000*√(t-2)
<span> </span>
Answer:
Coordinates: (0,0) ; (1,5) ; (2,10)
Step-by-step explanation:
x | y
0 0
1 5
2 10
Coordinates: (0,0) ; (1,5) ; (2,10)
This should be a straight line that is going diagonally with a positive slope.
Answer:
(9.5, 0) is in quadrant I. (-4, 7) is in quadrant II. (-1, -8) is in quadrant III.
Step-by-step explanation:
The negative signs say everything (quite literally). If there are no negative signs, it is in quadrant I. If there is one in the x-axis (the first number in an ordered pair), it is in quadrant II. If there are 2 negative signs, it is in quadrant III, and if there is one in the y-axis (the second number in an ordered pair), it is in quadrant IV.
