The point R(-3,a,-1) is the midpoint of the line segment jointing the points P(1,2,b) and Q(c,-7,4)
Find the values of:
a=
b=
C=

The values are:

a = -5/2

b = -6

c = -7

Step-by-step explanation:

Given

P = (x₁, y₁, z₁) = (1, 2, b)  

Q =  (x₂, y₂, z₂) = (c, -7, 4)  

As the point R(-3, a, -1) is the midpoint of the line segment jointing the points P(1,2,b)  and Q(c,-7,4), so

m = R = (x, y, z) = (-3, a, -1)

Using the mid-point formula

given

(x₁, y₁, z₁) = (1, 2, b) = P

(x₂, y₂, z₂) = (c, -7, 4) = Q

m = (x, y, z) = (-3, a, -1) = R

substituting the values in the mid-point formula

as (x, y, z) = (-3, a, -1), so

so solving 'c'

-3 = (1+c) / (2)

-3 × 2 = 1+c

1+c = -6

c = -6 - 1

c = -7

solving 'a'

a = (2+(-7)) / 2

2a = 2-7

2a = -5

a = -5/2

solving b

-1 = (b+4) / 2

-2 = b+4

b+4 = -2

b = -2-4

b = -6

Thus, the values are:

a = -5/2

b = -6

c = -7

Verification:

put a = -5/2, b = -6, c = -7

1.true 2.false? ? i think. .

step-by-step explanation:

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! : )

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