The "standard" parabola with roots 0 and 2 is
All multiples of this parabola, i.e.
have the same roots. We can choose the factor such that the parabola passes through the desided point: if we plug 1, 5 for x, y we have
So, our claim is that the parabola
has roots 0 and 2 and vertex at (1, 5).
You can easily verify this: the roots are guaranteed by the fact that we can write the equation as
The vertex must be at x=1, because it's the midpoint of the roots. Moreover, if we evaluate the function at x=1 we have
as required.
Answer:
Radius=3.67
Step-by-step explanation:
The screenshot shows the answer of the radius beng 3.67cm.
We need a system of equations here, one equation based on the NUMBER of tickets sold and another based on the MONEY earned by the sales. We have 2 different types of tickets: full price (f) and discount (d). The total number of tickets sold is 428; therefore, the first equation is f + d = 428. That accounts for the number of tickets sold. Each full price is 10.25 which can be represented as 10.25f, and each discount ticket costs 8 which can be represented as 8d. The money earned by selling these tickets at those prices was 3946. That means that the second equation is 10.25f + 8d = 3946. We will solve the first bolded equation for f to get f = 428 - d. Sub that value in for f in the second bolded equation: 10.25(428-d) + 8d = 3946. Distribute to get 4387 - 10.25d + 8d = 3946. Combine like terms to get -2.25d = -441. Solving for d we get 196. That means that there were 196 discounted tickets sold. Put that in for d in the first bolded equation to find the number of full price tickets. f + 196 = 428, and f = 232. There were 232 full price tickets sold. There you go!
Let the three items be M, Y and P.
n{M ∩ Y} only = 4-3 = 1
n{M ∩ P) only = 5-3 = 2
n{ Y ∩ P} only = 2
n{M} only = 12-(1+3+2) = 6
n{Y} only = 10-(1+2+3) = 4
n{P} only = 14-(2+3+2) = 7
n{M∩P∩Y} = 3
Number of women in the group = 6+4+7+(1+2+2+3) as above =25 women.