hectictar

Pick a state from the list.....for example, Nevada.

On a map, draw rectangles and triangles to roughly fill in the shape of the state, like this:

You're right that   \((\frac{2x-5}{5-x})(\frac{5-x}{5-x})\)  is equivalent to  \(\frac{-2x^2+15x-25}{x^2-10x+25}\)

....What do the full directions say?

1^2   =   1 * 1   =   1

Does that help? 

Let the coordinates of the third point be  (x, y) .  Since the triangle is equilateral...

distance between  (x, y)  and  (-1, 2)   =   distance between  (x, y)  and  (4, 2)

\(\sqrt{(-1-x)^2+(2-y)^2}\,=\,\sqrt{(4-x)^2+(2-y)^2} \\ (-1-x)^2+(2-y)^2\,=\,(4-x)^2+(2-y)^2 \\ (-1-x)^2\,=\,(4-x)^2 \\ (-1-x)\,=\,\pm(4-x)\\ \begin{array}\ (-1-x)=+(4-x)\qquad&\text{or}&\qquad(-1-x)=-(4-x)\\ -1-x=4-x&&\qquad-1-x=-4+x\\ -1=4&&\qquad-1=-4+2x\\ \text{not a solution}&&\qquad3=2x\\ &&\qquad\frac32=x \end{array}\)

So we know that the x coordinate must be  3/2 , which is 1.5 .

To find the y coordinate, we need to make another equation.

distance between  (-1,2)  and  (4, 2)   =   4 - -1   =   5      So...

distance between (-1, 2)  and  (1.5, y)   =   5

\(\sqrt{(-1-1.5)^2+(2-y)^2}=5\\ \sqrt{6.25+(2-y)^2}=5\\ 6.25+(2-y)^2=25\\(2-y)^2=18.75\\2-y=\pm\sqrt{18.75}\\ -y=\pm\sqrt{18.75}-2\\ y=\pm\sqrt{18.75}+2\\ \begin{array}\ y=\sqrt{18.75}+2\qquad\text{or}\qquad&&y=-\sqrt{18.75}+2\\ y\approx6.3&&y\approx-2.3 \end{array}\)

So the solutions for  (x, y)  are:  (1.5, 6.3)  and  (1.5, -2.3)

AB  =  \(\sqrt{(3-1)^2+(5-3)^2}\,=\,\sqrt{2^2+2^2}\,=\,\sqrt{4+4}\,=\,\sqrt8\)

BC  =  \(\sqrt{(3-1)^2+(3--1)^2}\,=\,\sqrt{2^2+4^2}\,=\,\sqrt{4+16}\,=\,\sqrt{20}\)

CD  =  \(\sqrt{(5-3)^2+(3--1)^2}\,=\,\sqrt{2^2+4^2}\,=\,\sqrt{4+16}\,=\,\sqrt{20}\)

DA  =  \(\sqrt{(5-3)^2+(5-3)^2}\,=\,\sqrt{2^2+2^2}\,=\,\sqrt{4+4}\,=\,\sqrt8\)

AB + BC + CD + DA   =   √8 + √20 + √20 + √8   ≈  14.6

avg rate of change over the interval [5, 7]   =   \(\frac{f(7)-f(5)}{7-5}\)

avg rate of change over the interval [5, 7]   =   \(\frac{1469-549}{7-5}\)

avg rate of change over the interval [5, 7]   =   \(\frac{920}{2}\)

avg rate of change over the interval [5, 7]   =   460

avg rate of change over the interval [2, 4]   =   \(\frac{f(4)-f(2)}{4-2}\)

avg rate of change over the interval [2, 4]   =   \(\frac{287-39}{4-2}\)

avg rate of change over the interval [2, 4]   =   \(\frac{248}{2}\)

avg rate of change over the interval [2, 4]   =   124

How much greater is the avg rate of change over the interval  [5, 7]  than the avg rate of change over the interval  [2, 4] ? That is...how much greater is  460  than  124 ?

460 - 124  =  336

460  is  336  greater than  124 .

Find  \(a+b+c\) , given that  \(x+y\neq -1\)  and  \( \begin{align*} ax+by+c&=x+7,\\ a+bx+cy&=2x+6y,\\ ay+b+cx&=4x+y. \end{align*}\)

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(ax + by + c) + (a + bx + cy) + (ay + b + cx)  =  (ax + by + c) + (a + bx + cy) + (ay + b + cx)

(ax + by + c) + (a + bx + cy) + (ay + b + cx)  =  (x + 7) + (2x + 6y) + (4x + y)

ax + by + c + a + bx + cy + ay + b + cx  =  x + 7 + 2x + 6y + 4x + y

ax + bx + cx + ay + by + cy + a + b + c  =  7x + 7y + 7

x(a + b + c) + y(a + b + c) + 1(a + b + c)  =  7(x + y + 1)

(a + b + c)(x + y + 1)  =  7( x + y + 1)

a + b + c  =  7

(f + g)(x)   =   f(x) + g(x)             Since  f(x)  =  x³ - x ,  we can replace  f(x)  with  x³ - x .

(f + g)(x)   =   (x³ - x) + g(x)        Since  g(x)  =  x³ + 2x² - 10 ,  we can replace  g(x)  with  x³ + 2x² - 10 .

(f + g)(x)   =   (x³ - x) + (x³ + 2x² - 10)        We can drop the parenthesees.

(f + g)(x)   =   x³ - x + x³ + 2x² - 10         Addition can be done in any order.

(f + g)(x)   =   x³ + x³ + 2x² - x - 10         Combine like terms.

(f + g)(x)   =   2x³ + 2x² - x - 10