MathProblemSolver101

This is correct. Nice job!

Perhaps, you notice in your answer that \(2115 = 46^2 - 1\)? This is no coincidence.

Instead of letting \(f(0)=0\), we let \(f(0)=1\)

And it follows that \(\sum_{i=0}^{99}f(i)=(1+1+2+3+\cdots+9)^2=46^2\)

Subtracting \(f(0)\), we have 

\(\begin{align*} 46^2 - f(0) &= 46^2 - 1 \\ &= 2116 - 1 \\ &= \boxed{2115} \end{align*}\)

XZ = sqrt(28^2 - 24^2) = 4 sqrt 13

tan y = 4 sqrt 13 / 24 = sqrt 13 / 6

62 + ... + 162

2(31 + ... + 81)

2((1 + ... + 51) + 51 * 30)

2(1 + ... + 51) + 3060

51*52 + 3060

5712

x*xsqrt3*1/2 = 2x*3*1/2

x^2 sqrt3 = 6x

xsqrt3=6

x=2sqrt3

(1) we have the altitude = 20 so 20*30*1/2 = 300

(2) 13/2 = 6.5

note 1/a + 1/b = (a+b)/ab

a+b = 11/5

ab = 4/5

(11/5)/(4/5) = 11/4

Note that Vieta's can be generalized for degree $n.$ Try finding a formula.

\(ax^2 + bx + c = a(x-r_1)(x-r_2) \\ ax^2 + bx + c = a(x^2 + x(-r_1 - r_2) + r_1 r_2) \\ ax^2 + bx + c = a \cdot x^2 + a \cdot x(-r_1 - r_2) + a \cdot r_1 r_2 \\ a(-r_1-r_2)=b \\ (-1)(a)(r_1+r_2)=b \\ r_1 + r_2 = - \frac{b}{a} \\ a \cdot r_1 r_2 = c \\ r_1 r_2 = \frac{c}{a}\)

Hi Mobius,

You are correct. There is a second answer to the question, but does it fit in the range?

Best,

MPS

$\sec(x) = \frac{1}{\cos(x)}$

\(\begin{align*} \tan^2(x) &= \left(\frac{\sin x}{\cos x} \right)^2 \\ &= \frac{\sin^2 x}{\cos^2 x} \\ &= \frac{\sin^2 x}{\cos^2 x} + \frac{\cos^2 x}{\cos^2 x} - 1\\ &= \frac{\sin^2 x + \cos^2 x}{\cos^2 x} - 1\\ &= \frac{1}{\cos^2 x} - 1 \end{align*}\)

$\frac{5}{\cos(x)} = \frac{3}{\cos^2 x} - 2$

$5 \cos(x) = 3 - 2 \cos^2 x$

$2 \cos^2 x + 5 \cos x - 3 = 0$

$(\cos x + 3) (2 \cos x - 1) = 0$

$\cos x = -3$

$\cos x = \frac{1}{2}$

I think you can take it from here.