CPhill

No way to do this...we need to know at least one of the acute angles, as well.....

H = 50 tan 28 = about 26.6 ft

Let f(x)=-3x^2+kx+24,where k is a constant .If the y-coordinate of the vertex is 27, then k =??

Taking the derivative, we have

f'(x) = -6x + k ....and at the vertex, this = 0 so we have..

-6x + k  = 0     which implies that k = 6x at this point   .....so we have

27 = -3x^2 + (6x)x + 24  = -3x^2 + 6x^2 + 24   simplify

3 = 3x^2 

1 = x^2  →  x = ± 1

So  when x = ± 1,  then k = ±6

When x = 1, k = 6 and we have

27 = -3(1)^2 +6(1) + 24   which is true

And when x = -1, k = -6

27 =-3(-1)^2 -6(-1) + 24  which is also true

15x-12y=13....this is just an equation of a line...it would have infinite solutions for x and y......

Here's a graph.....

C = 3.14 x diameter

So, when d = 4, C = 12.56

I assume that you would have finished earlier, Melody,  had you not been.... "HEXED"......!!!!

I assume that the chemist is mixing the first two to get the 20L of 60% solution.....we need to find out how many L of each of the first two must be combined to produce such a mixture.

So we have

.15x + .90(20-x) = 20(.60)

.15x + 18 - .90x = 12   simplify

6 = .75x   → x = 8 L of 15% solution

So (20-8) = 12 L of the 90% solution were used

Now......5L of this is poured back into the the bottle of 90% solution that now contains (30 - 12) = 18L

So we have

5L of 60% HCL = 3 L of HCL  ...  and 18L of 90% HCL  = 16.2 L of HCL

Then

(3 + 16.2) L of HCL / ( 5+ 18) total L of solution  = about an 83.5% concentration of HCL....

Here's another way to do this just by equating coefficients.....

a(x^2 - 2px  + p^2) + q  = -(1/2)x^2 + 5x + 20

ax^2 - 2apx + p^2 + q   =  -(1/2)x^2 + 5x + 20

Equating coefficients 

a = -(1/2)

-2ap = 5 →   (-2)(-1/2)p  = 5 → p  = 5

And

ap^2 + q = 20

(-1/2)(5)^2 + q = 20  →   -25/2 + q = 20 →  q = 20 + 25/2  = 65/2 = 32.5 =  q

The vertex is equidistant from the focus and the directrix......so in this case, the vertex is just [(1+7)/2 , 3) =

(4, 3)....BTW...this parabola opens to the left.....