The sixth grade runs a bake sale for 55 hours and makes $170.
So in one hour, they made 170/55 = 3.09 dollars.
Rate: $3.09/h
The seventh grade sets up a dunking booth for 44 hours and makes $112.
So in one hour, they made 112/44 = 2.54 dollars.
Rate: $2.54/h
The eight grade has a car wash and makes $192 in 66 hours.
So in one hour, they made 192/66 = 2.90 dollars.
Rate $2.90/h
3.09 > 2.90 > 2.54
So R(6th) > R(8th) > R(7th)
Which means that the sixth grade has the highest rate for raising money.
Hope this helps! :)
Answer:
q = -8, k = 2.
r = -6.
Step-by-step explanation:
f(x) = (x - p)^2 + q
This is the vertex form of a quadratic where the vertex is at the point (p, q).
Now the x intercepts are at -6 and 2 and the curve is symmetrical about the line x = p.
The value of p is the midpoint of -6 and 2 which is (-6+2) / 2 = -2.
So we have:
f(x) = 1/2(x - -2)^2 + q
f(x) = 1/2(x + 2)^2 + q
Now the graph passes through the point (2, 0) , where it intersects the x axis, therefore, substituting x = 2 and f(x) = 0:
0 = 1/2(2 + 2)^2 + q
0 = 1/2*16 + q
0 = 8 + q
q = -8.
Now convert this to standard form to find k:
f(x) = 1/2(x + 2)^2 - 8
f(x) = 1/2(x^2 + 4x + 4) - 8
f(x) = 1/2x^2 + 2x + 2 - 8
f(x) = 1/2x^2 + 2x - 6
So k = 2.
The r is the y coordinate when x = 0.
so r = 1/2(0+2)^2 - 8
= -6.
Answer:
b= 29/11
Step-by-step explanation:
2b+8−5b+3=−13+8b−5
Step 1: Simplify both sides of the equation.
2b+8−5b+3=−13+8b−5
2b+8+−5b+3=−13+8b+−5
(2b+−5b)+(8+3)=(8b)+(−13+−5)(Combine Like Terms)
−3b+11=8b+−18
−3b+11=8b−18
Step 2: Subtract 8b from both sides.
−3b+11−8b=8b−18−8b
−11b+11=−18
Step 3: Subtract 11 from both sides.
−11b+11−11=−18−11
−11b=−29
Step 4: Divide both sides by -11.
−11b/−11 = −29/−11
b= 29/11
Ok for the last time stop cheating and the answer is b
