You have to add the same kind of terms, in this case the ones that have a variable (5.4q and -4.5q) and the ones that don't (2.4 and 3.6), like this:
(5.4q - 4.5q) + (2.4 + 3.6) = 0.9q + 6
The answer is c. 0.9 +6
Substitute y = -x into the first equation:
-8x - 7(-x) = -5
-8x + 7x = -5
-x = -5
x = 5
Substitute x = 5 into y = -x
y = -(5)
y = -5
Answer: x = 5, y = -5
Here, first we need to calculate the slope of the line,
m = y2 - y1 / x2-x1
m = 6 + 2 / 3 -1
m = 8/ 2
m = 4
Now, Take first coordinate: y - 6 = 4 (x - 3)
Second coordinate: y + 2 = 4 (x - 1)
In short, Your Answers would be Option A & F
Hope this helps!
Step-by-step explanation:
This is the answer.... Working shown
Speed of the plane: 250 mph
Speed of the wind: 50 mph
Explanation:
Let p = the speed of the plane
and w = the speed of the wind
It takes the plane 3 hours to go 600 miles when against the headwind and 2 hours to go 600 miles with the headwind. So we set up a system of equations.
600
m
i
3
h
r
=
p
−
w
600
m
i
2
h
r
=
p
+
w
Solving for the left sides we get:
200mph = p - w
300mph = p + w
Now solve for one variable in either equation. I'll solve for x in the first equation:
200mph = p - w
Add w to both sides:
p = 200mph + w
Now we can substitute the x that we found in the first equation into the second equation so we can solve for w:
300mph = (200mph + w) + w
Combine like terms:
300mph = 200mph + 2w
Subtract 200mph on both sides:
100mph = 2w
Divide by 2:
50mph = w
So the speed of the wind is 50mph.
Now plug the value we just found back in to either equation to find the speed of the plane, I'll plug it into the first equation:
200mph = p - 50mph
Add 50mph on both sides:
250mph = p
So the speed of the plane in still air is 250mph.
