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wolverine [178]
4 years ago
5

A polynomial function has roots –5 and 1. Which of the following could represent this function? f(x) = (x + 5)(x + 1) f(x) = (x

– 5)(x – 1) f(x) = (x – 5)(x + 1) f(x) = (x + 5)(x – 1)
Mathematics
2 answers:
jarptica [38.1K]4 years ago
6 0

Answer:

2nd part of the question:  The corresponding polynomial function is:

F(X)=X^2+4X-5


Step-by-step explanation:


Dimas [21]4 years ago
3 0

f(x) = (x + 5)(x - 1)

using the ' factor theorem '

given x = a is the root of a polynomial then (x - a ) is a factor

here roots are x = - 5 and x = 1 hence factors are (x + 5) and (x - 1)

the polynomial is the product of the factors

f(x) = (x + 5)(x - 1)


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8 0
3 years ago
You have a coupon for $5 off your total bill at Mama's Meals on Main a. How much will you pay after using the coupon if your bil
HACTEHA [7]

Answer :

(a) Amount paid after using the coupon = $b - $5 = $(b-5)

(b) Amount paid = $23.45 - $5 = $(23.45-5) = $18.45

Amount paid = $54.83- $5 = $(54.83-5) = $49.83

Step-by-step explanation :

As we are given that:

Amount of coupon = $5

If amount of total bill = $b

Now we have to calculate the amount paid after using the coupon.

Amount paid after using the coupon = Total amount of bill - Amount of coupon

Amount paid after using the coupon = $b - $5 = $(b-5)

Now we have to calculate the amount paid if bill was $23.45.

Amount paid after using the coupon = Total amount of bill - Amount of coupon

Amount paid = $23.45 - $5 = $(23.45-5) = $18.45

Now we have to calculate the amount paid if bill was $54.83.

Amount paid after using the coupon = Total amount of bill - Amount of coupon

Amount paid = $54.83- $5 = $(54.83-5) = $49.83

6 0
4 years ago
Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)
DENIUS [597]

Compute the necessary values/derivatives of f(x) at x=1:

f(1)=3

f'(1)=2

f''(1)=-12

f'''(1)=-12

f^{(4)}(1)=48

f^{(5)}(1)=120

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) f(x) at x=1 by

T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5

T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5

###

Another way of doing this would be to solve for the coefficients a,b,c,d,e,g in

f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5

by expanding the right hand side and matching up terms with the same power of x.

5 0
3 years ago
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