Consider the equation a = -2b² + 1. Determine the graph of the relationship between a and b. Choice A
Choice B
Choice C
Choice D

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explanation

the graph has a slope of 1/3 because each time you move up 1, you go to the right 3. so, slope = rise/run = 1/3. we can cross choice a off the list since we want a slope of 3, rather than 1/3.

the first table shown has a slope of -3. this is because the rise is -6 and the run is 2, so slope = rise/run = -6/2 = -3. note how x increases by 2 each time (run) and y drops by 6 each time (rise). choice b can be eliminated.

the second table has a slope of 3. each time x goes up by 1, the y values increase by 3. slope = rise/run = 3/1 = 3. alternatively you can use the slope formula to prove that the slope is 3 here. use any two rows of values to generate the two points. choice c is one of the answers

the third table also has a slope of 3. let's use the slope formula to prove as such. i'm going to use the first two rows of the table. so i'll use the two points (2,1) and (5,10)

m = slope

m = (y2-y1)/(x2-x1)

m = (10-1)/(5-2)

m = 9/3

m = 3 so we get a slope of 3

choice d is the other answer

step-by-step explanation:

descending order means to order the polynomials from the term with the largest exponent to the term with the smallest exponent.  

therefore, the correct answer is d.  

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

d

step-by-step explanation:

i used the pitagoras theoreme