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Aliun [14]
3 years ago
10

Help help help help help

Mathematics
1 answer:
Aleks [24]3 years ago
5 0

Answer:

I think it is A ..... ........

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Jamison graphs the function ƒ(x) = x4 − x3 − 19x2 − x − 20 and sees two zeros: −4 and 5. Since this is a polynomial of degree 4
Art [367]

We are given polynomial function f(x)=x^4-x^3-19x^2-x-20

We are given two zeros -4 and 5.

Therefore, x=-4 and x=5 is given.

So, the two factors of the polynomial will be (x+4)(x-5).

Let us write some steps to factor the given polynomial.

Let us take above factor (x+4) first.

On factoring (x+4) we get factors

x^4-x^3-19x^2-x-20=\left(x+4\right)\left(x^3-5x^2+x-5\right)

\mathrm{:Let \ us \ factor}\:x^3-5x^2+x-5 \ now.

Grouping

=\left(x^3-5x^2\right)+\left(x-5\right)

Factoring out gcf from each group, we get

=x^2\left(x-5\right)+\left(x-5\right)

=\left(x-5\right)\left(x^2+1\right)

So, the final factored form of given polynomial will be

x^4-x^3-19x^2-x-20=\left(x+4\right)\left(x-5\right)\left(x^2+1\right)

For first to factors (x+4) and (x-5) we have given roots: -4 and 5.

Let us find the root of third factor we got.

x^2+1=0

Subtracting both sides by 1.

x^2 = - 1.

On taking square root on both sides, we get a square root(-1)

x=\sqrt{-1}=+ i and -i.

Those are imaginary solutions.

Therefore, correct option is B) No, there are two imaginary solutions.



5 0
4 years ago
What’s in between 10.3 and 12.1
Brums [2.3K]

Answer:

11.3

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
PLEASE HELP a three dight number has one more ten than it has hundreds, and it also has one more than twice as many units as ten
bija089 [108]

The reverse number of the three-digit number is 732

<h3>How to determine the reverse of the number?</h3>

Let the three-digit number be xyz.

So, the reverse is zyx

This means that

Number = 100x + 10y + z

Reverse = 100z + 10y + x


From the question, we have the following parameters:

y = x + 1

z = 1 + 2y

The sum is represented as:

100x + 10y + z + 100z + 10y + x = 10^3 - 31

100x + 10y + z + 100z + 10y + x = 969

Evaluate the like terms

101x + 101z + 20y = 969

Substitute y = x + 1

101x + 101z + 20(x + 1) = 969

101x + 101z + 20x + 20 = 969

Evaluate the like terms

101x + 101z + 20x = 949

121x + 101z = 949

Substitute y = x + 1 in z = 1 + 2y

z = 1 + 2(x + 1)

This gives

z = 2x + 3

So, we have:

121x + 101z = 949

121x + 101* (2x + 3) = 949

This gives

121x + 202x + 303 = 949

Evaluate the sum

323x = 646

Divide by 323

x = 2

Substitute x = 2 in z = 2x + 3 and y = x + 1

z = 2*2 + 3 = 7

y = 2 + 1 = 3

So, we have

x = 2

y = 3

z = 7

Recall that

Reverse = 100z + 10y + x

This gives

Reverse = 100*7 + 10*3 + 2

Evaluate

Reverse = 732

Hence, the reverse number of the three-digit number is 732

Read more about digits at:

brainly.com/question/731916

#SPJ1

8 0
1 year ago
A bag contains 5 blue marbles, 6 red marbles, and 9 green marbles. 2 marbles are picked at random. What is the probability of fi
taurus [48]
Its a 50%-50% chance that you will be able to pick a red and blue marble without noticing....

8 0
3 years ago
What is 48+6x factorised
mafiozo [28]

Answer:

54x

Step-by-step explanation:

48+6=54 add the x and your answer is 54x

7 0
3 years ago
Read 2 more answers
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