Step-by-step explanation:
x in (-oo:+oo)
1/7-(3*((3/7)*x-(2/7))) = 0
1/7-3*((3/7)*x-2/7) = 0
1/7-3*(3/7*x-2/7) = 0
(-3*7*(3/7*x-2/7))/7+1/7 = 0
1-3*7*(3/7*x-2/7) = 0
7-9*x = 0
(7-9*x)/7 = 0
(7-9*x)/7 = 0 // * 7
7-9*x = 0
7-9*x = 0 // - 7
-9*x = -7 // : -9
x = -7/(-9)
x = 7/9
x = 7/9
<span>Let n = the number of nickles
Let q = the number of quarters
Then for your problem we have
(1) n + q = 43 and
(2) 5*n + 25*q = 100*6.95 (always work in cents to avoid decimal numbers) or
(3) 5*n + 25*q = 695
Now substitute n of (1) into (3) and get
(4) 5*(43 - q) + 25*q = 695 or
(5) 215 - 5*q + 25*q = 695 or
(6) 20*q = 695 - 215 or
(7) 20*q = 480 or
(8) q = 24
Then using (1) we get
(9) n + 24 = 43 or
(10) n = 19
Let's check these values.
Is (.05*19 + .25*24 = 6.95)?
Is (.95 + 6.00 = 6.95)?
Is (6.95 = 6.95)? Yes
Answer: Kevin and Randy have 19 nickles and 24 quarters in the jar.</span>
f (x) = a(x - h)^2 + k, where (h, k) is the vertex of the parabola.
Do you see the h and k in your equation?
-h = -(-2) = 2
We see that k = -4.
Our vertex is (2, -4).
Ok so first you will need 3 bags since 12*2 is 24 that wont be enough for the whole class so 12*3= 36 so now that we figured out how many bags we need now we have to multiply $8.00 by three the answer is $24
