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I couldn't see a neat way of doing this analytically, so I used a numerical approach:

Let's first simpify the quadratic. Moving all terms to the left hand side, we get \(2x^2-2x-10=0 \). We want to find the value of \(a+b+6\). Instead of finding the roots of the quadratic, we can use Vieta's Formula. So the sum of the roots would be \(-\left(-\dfrac{2}{2}\right)\). \(1+6=\boxed{7}\). And now we're done!  Hope this helped!

Let's first rearrange the equation to the standord form. Moving all terms to the right, we get \(19x^2-6x+31=0 \). We apply the quadratic formula and we get the roots are \(3 \pm 25i\)  So \(a\) and \(b\) are \(3\) and \(25\). \(3 \cdot 25=\boxed{75} \). And now, we're done!  Hope this helped!

Try writing a in terms of b or b in terms of a to solve the equation. Like: a = b + 1. Then just replace it: a^3 - b^3 = (b-1)^3 - b^3 = 1. Expand, simplify, solve. 

we need:
\(f(f(x))=x\\ f(x)=f^{-1}(x)\)
The only way this is possible is if \(f(x)=f^{-1}(x)=x\):
So:
\(\cfrac{cx}{(4x-5)(3x+2)}=x\\ \boxed{\displaystyle{c=(4x-5)(3x+2)}}\)

Hello blackpanther!

I have corrected AX on PX.

\(\frac{\angle YPM}{2}=atan\ \frac{8}{18}=23.962^\circ\\ \angle RPM=47.925^\circ\\ h_{via\ PQ}=22\cdot sin\ \angle RPM=22\cdot sin\ 47.925^\circ\\ h_{via PQ}=16.32\\ A_{\triangle PQR}=\frac{1}{2}\cdot 36\cdot 16.32\\ \color{blue}A_{\triangle PQR}=293.76\)

 !

OCF NPV = $249,000 + 18,000 − [$115,000(1 − .21) + $18,000]/1.143
OCF NPV = $193,529.35

$193,529.35 = OCF{[1 − (1/1.143)]/.14}
OCF = $83,359.18

NI = $83,359.18 − $249,000/3
NI = $359.18

EBT = $359.18/(1 − .21)
EBT = $454.66

Sales = $454.66 + 37,000 + $249,000/3 + $148,000(2)
Sales = $416,454.66

Bid per park = $416,454.66/2
Bid per park = $208,227.33 or $208,200

We know, \(\sin(x)=\cos(\frac{\pi}{2}-x)\).

\(\angle BAD = 90 = \frac{\pi}{2}\), exactly!

\(\because \angle BAC = \angle BAP + \angle PAQ + \angle QAD = \frac{\pi}{2}\)

\(\therefore\angle BAP+\angle QAD=\frac{\pi}{2}-\angle PAQ\)

\(\therefore \sin(\angle PAQ)=\cos(\angle BAP + \angle QAD)=\cos(\angle BAP)\cos(\angle QAD)-\sin(\angle BAP)\sin(\angle QAD)\).

Triangle BAP and QAD are just right triangles, so it's easy to calculate trigonometric values off them.

We get \(\frac{17\sqrt{1189}}{1189}\), as our final answer.

The rectangle with these coordinates is centered at the origin, with side length 2. Note that any equation y = mx will bisect the area of the square, so the y-intercept is 0. (Try drawing it on graph paper too) 

f(x) = a{trig}(bx - c) + d

a is the amplitude

2pi/|b| = the period

c/b = the horizontal phase shift

d = midline

Following this generalization, the midline is y = 1.8

0.03 + 0.0014 repeating = 0.0314 repeating

Let 0.0014repeating = S, 100S = 0.141414repeating = S + 0.14

Solving, you have 99S = 0.14, S = 0.14/99 = 14/(99)(100) = 14/9900

0.03 + 7/4950 = 297/9900 + 14/9900 = 311/9900

A cubic polynomial, by the fundamental theorem of algebra, will have EXACTLY three roots (which generalizes to an nth degree polynomial to have n complex roots). Clearly, in this problme, there are four roots, so a cubic polynomial f does not exist, and all coefficients would be 0. Thus, the sum of the coefficients is 0.

We know:

\(\cos(2x)=1-2 {\sin}^{2}(x)=2{\cos}^{2}(x)-1\).

Using a clever implementation of these formulas,

\(\cos(4x)=2{\cos}^{2}(2x)-1=2{(1-2{\sin}^{2}(x))}^{2}-1\).

We get: \(\cos(4x)=8{\sin}^{4}(x)-8{\sin}^{2}(x)+1\). 

We have a very interesting property, and very useful property when it comes to slope: \(\tan=\frac{O}{A}\), opposite over adjacent, but when put in the coordinate plane, that is \(\tan=\frac{rise}{run}\), and what else is rise over run? Slope! .

Set the angle formed by y=5x and the x axis as \(\theta\), and the angle formed by y=1/6x and the x axis as \(\phi\).

We want to find \(\tan(\phi+\frac{\theta-\phi}{2})=\tan(\frac{\theta+\phi}{2})\).

Use some of our identities:

With a lot of use of tangent identities, we know \(\tan(\theta)=5, \tan(\phi)=1/6\).

Directly applying our tangent angle identity: \(\tan(\theta+\phi)=\frac{5+\frac{1}{6}}{1-\frac{5}{6}}=31\).

Calculate, \(\tan(\theta+\phi)=\frac{\sin(\theta+\phi)}{1+\cos(\theta+\phi)}\).

Draw a triangle and calculate sin and cos.

\(\sin(\theta+\phi)=\frac{31}{\sqrt{962}}\), \(\cos(\theta+\phi)=\frac{1}{\sqrt{962}}\)

Substituting in, we get \(y=(\frac{\sqrt{962}-1}{31})x\) as our line, so \(\bf{m=\frac{\sqrt{962}-1}{31}}\).\(\)

If you draw a diagram, and calculate, you will get \(\boxed{\dfrac{2}{3}}\)

To find the area of a trapazoid, we need the find the height and the average of the bases of the trapazoid. The height of the trapazoid is \(16\). The average of the bases is \(25\). \(\dfrac{25 \cdot 16}{2}=200 \). And now, we're done!  Hope this helped!

By testing rows of Pascal's triangle, we get row \(3\) satisfies the conditions above! So \(n\) is equal to \(3\). And now we're done!  Hope this helped!

There is only one way to arrange the three yellow squares into the five slots. Now we only have two more slots. Both slots each have \(2\) choices for a color: red or blue. So \(2 \cdot 2 = \boxed{4}\)

That is the number of ways to color the squares. And now, we're done!  Hope this helped!