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Try thinking about a different question.

find the roots of    \(y=x^2+5x+6\)

To do this (without a formula) I must first factorize it.

\(y=(x+2)(x+3)\)

Now if x=-2 then y will be 0

and if x=-3 then y will also be 0.

So

x=-3 and x=-2 are roots of this equation.

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Now I will go in the other direction:

If  x=-3 and x=-2 are roots of     \(y=x^2+5x+c\)     find c.

I would say

y=(x--3)(x--2)

y=(x+3)(x+2)

\(y=x^2+5x+6\)

so     c=6.

Think about it  

The largest possible degree of f(x) + a*g(x) is 3.

I am still relatively new to trig, what is K?

sin law 

16 is to sin 51     as     BC is to sin 84

16 / sin51    =    BC / sin 84     

sin 84 *   16 / sin51 = BC

Start with a sin graph    (you could also use a cos graph)

change amplitude from 1    to  6   (the height of the wave is  -2 to -14  a distance of 12 ....amplitude is 1/2 height)

6 sin x       

shift the midline down from 0 to -8

6 sin x    - 8

change period from 2pi   to 2.5 pi       2pi / 2.5 pi = .8      Multiply the object of the sin fxn by .8

6 sin (.8x )    - 8

shift peak from  the new peak    5pi/8    RIGHT to 6 pi/8     so subtract  pi/8 from x

6 sin .8(x - pi/8) - 8           simplify a bit

6 sin (.8x - .1pi) - 8

Here is a graph :  

https://www.desmos.com/calculator/lpxwuenxzi

That one angle at the tangent is marked 74^o  

So the angle at the other tangent is also 74^o  

Together those two angles total 148^o   

The total of the internal angles of a triangle is 180^o  

So angle m equals 180^o minus 148^o       m  =  32^o   

_.

(77.25   +  77.5    +  81   +  x  ) /  4    = 80        solve for 'x'        Sorry....typo

hrm, those are both incorrect... 

No worries... :)

I've done these problems!

1. Angle YXZ = 46 degrees.

2. The area of trapezoid EFGH is 50.

(77.25   +  77.5    +  81   +  x  ) /  5    = 80        solve for 'x'

The expected value is 15/4.

FG is   1 / (1+6) = 1/7 of the way from F

x:    From  4 to -8 is -12 units       1/7 * (-12)  added to 4  = .........

y:   From -5 to 19 is 24 units        1/7 * (24)  added to -5 = ........

The quadratic is f(x) = 2x^2 + 3x - 1.

100a + 10b + c = 148.

The largest triangle, by area, that can be inscribed in a triangle is the equilateral triangle.  Since you need a right triangle, you can swing one of the points to the other, which gives you a 30-60-90 triangle, with sides 50, 50*sqrt(30), and 100.  The area of this triangle is 1/2*50*50*sqrt(3) = 1250*sqrt(3).

In the triangle, you can write R = (abc)/(4K).  You should then be able to work out the rest.

f(g(f(2)) = f(g(11)) = f(36) = 111.

We can rewrite 4^(x-1) as 4^x/4. 4^x can be rewritten as (2^x)^2, which is 7^2, which is 49. So, the answer is 49/4, which is 12.25 in decimal form.

:)

Perfect squares we can add together    0   1    4     9    16

Sum of two perfect squares can total       1     4      9     16      5    10   17   13       eight total <19

At x = 0 , 1 2 3 4 5 6 7 8 9      the 'y' coordinate will be an integer >= 0

I'm confused at what a unique polygon is......you mean a shape that follows the "unique shape theorem"? Then it would be B, because a square is a regular rectangle, since all sides are equal. 
 

(Also, can you please reply, telling me what a unique polygon is? Thanks!)

:)

To start, let's find what a and b are. Using the quadratic formula, we see that a=(23+sqrt341)/2, and b=(23-sqrt341)/2. Now, plug that into what we want: ((23+sqrt341)/2-2)((23-sqrt341)/2-2). Multiply it out, and you get: 5.

:)

yeah...i find that when i use the latex box majority of the times it comes out as the raw form...help! 

All of the letters are different...5 choices

5 choices for first position

4 choices for second position

3 choices for third position

2 choices for fourth position

1 choice for last position

5 x 4 x 3 x 2 x 1 = 120