January . . . $57.85
March . . . . 4 times as much = 4 (57.85) = $231.40
Deposit 78.45 more . . . ($231.40 + 78.45) = <em>$309.85</em> .
Notice that "interest" is never mentioned anywhere in this problem.
In other words, it doesn't matter whether Julie's savings account is
in a bucket in the basement, a mayonnaise jar on the porch, under
her mattress, or in a bank that pays no interest.
Without interest, $309.85 is what she <em><u>does</u></em> have<em><u /></em> in November, which
is about right for savings accounts in banks these days.
What her balance <em><u>should</u></em> be in November is an entirely different subject.
Answer:
<h3>
(2, 124)</h3>
Step-by-step explanation:
f(x) = a(x - h)² + k - the vertex form of the equation of the parabola with vertex (h, k)
![f(x) = -16x^2+ 64x + 80\\\\f(x) = -16(x^2- 4x) + 80\\\\f(x) = -16(\underline {x^2-2\cdot2x\cdot2+2^2}-2^2) + 80\\\\f(x) = -16\big[(x-2)^2-4\big] + 80\\\\f(x) = -16(x-2)^2+64 + 80\\\\\bold{f(x)=-16(x-2)^2+124\quad\implies\quad h=2\,,\quad k=124}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20-16x%5E2%2B%2064x%20%2B%2080%5C%5C%5C%5Cf%28x%29%20%3D%20-16%28x%5E2-%204x%29%20%2B%2080%5C%5C%5C%5Cf%28x%29%20%3D%20-16%28%5Cunderline%20%7Bx%5E2-2%5Ccdot2x%5Ccdot2%2B2%5E2%7D-2%5E2%29%20%2B%2080%5C%5C%5C%5Cf%28x%29%20%3D%20-16%5Cbig%5B%28x-2%29%5E2-4%5Cbig%5D%20%2B%2080%5C%5C%5C%5Cf%28x%29%20%3D%20-16%28x-2%29%5E2%2B64%20%2B%2080%5C%5C%5C%5C%5Cbold%7Bf%28x%29%3D-16%28x-2%29%5E2%2B124%5Cquad%5Cimplies%5Cquad%20h%3D2%5C%2C%2C%5Cquad%20k%3D124%7D)
<u>The vertex is </u><u>(2, 124)</u>
AE x BE = DE x CE
6x5 = 30
6 x DE = 30
DE = 30/6 =5
DC = DE +CE = 5+6 =11
Answer is C) 11
