Answer:
x=36
Step-by-step explanation:
if PQR is isosceles then
PRQ will also equal 69 degrees
straight line = 180 degrees
so if PRQ is 69, then
PRS = 180-69
= 111 degrees
now to find x , since we have two angles within that triangle
180 - 111 - 33 = 36 degrees
21+
15
2
+
3
2
+
15
13
LCD = 1515
21+
15
2
+
3×5
2×5
+
15
13
21+
15
2+10+13
21+1
15
10
22
15
10
15
10
3
2
22
3
2
done
the answer of this question is last one
To solve/simplify this all you have to do is group like terms (the x^2's with each other, the x's with each other, and the normal numbers, -8)
14x^2-8+5x-6x^2+2x
group the x^2 (add 14x^2 to -6x^2)
8x^2-8+5x+2x
group the x's together (add 5x and 2x together)
8x^2+7x-8
Your answer will be d) 8x^2+7x-8
Answer:
there doesnt seem to be much information, but from what it looks like it looks like it is 20?
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.
