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Elden [556K]
3 years ago
10

Find all the zeros of the polynomial function p(x) = x3 – 5x2 + 33x – 29

Mathematics
1 answer:
hram777 [196]3 years ago
5 0

Answer:

\large \boxed{\sf \ \ x=1, \ \ x=2+5i, \ \ x=2-5i \ \ }

Step-by-step explanation:

Hello,

I assume that we are working in \mathbb{C}, otherwise there is only one zero which is 1. Please consider the following.

First of all, <u>we can notice that 1 is a trivial solution</u> as

   p(1) = 1^3-5\cdot 1^2 + 33\cdot 1-29=1-5+33-29=0

It means that (x-1) is a factor of p(x) so we can find two real numbers, a and b, so that we can write the following.

p(x)=(x-1)(x^2+ax+b)=x^3+ax^2+bx-x^2-ax-b=x^3+(a-1)x^2+(b-a)x-b

Let's identify like terms as below.

a-1 = -5 <=> a = -5 + 1 = -4

b-a = 33

-b = -29 <=> b = 29

So

\boxed{ \ p(x)=(x-1)(x^2-4x+29) \ }

Now, we need to find the zeroes of the second factor, meaning finding x so that:

x^2-4x+29=0 \ \text{ complete the square, 29 = 25 + 4}  \\ \\  x^2-2\cdot 2 \cdot x+2^2+25=0 \\ \\ (x-2)^2=-25=(5i)^2 \ \text{ take the root } \\ \\x-2=\pm 5i \ \text{ add 2 } \\ \\  x = 2+5i \ \text{ or } \ x = 2-5i

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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NEED HELP ASAP!!!<br> thank you.
Alex17521 [72]

Answer:

See explanation

Step-by-step explanation:

Consider triangles ACM and BCM. In these triangles,

  • m\angle 3=m\angle 4 - given;
  • m\angle 1=m\angle 2=90^{\circ} - definition of perpendicular lines CM and AB;
  • \overline{CM}\cong \overline{CM} - reflexive property.

So,

\triangle ACM\cong \triangle BCM by ASA postulate (if one side and two angles adjacent to this side of one triangle are congruent to one side and two angles adjacent to this side of another triangle, then two triangles are congruent).

Two-column proof:

      Statement                                 Reason

1. m\angle 3=m\angle 4                          Given

2. CM\perp AB                        Given

3. m\angle 1=m\angle 2=90^{\circ}       Definition of perpendicular lines CM and AB

4. \overline{CM}\cong \overline{CM}          Reflexive property

5. \triangle ACM\cong \triangle BCM                       ASA postulate

7 0
2 years ago
A rectangle has a length of the fifth root of 16 inches and a width of 2 to the 1 over 5 power inches. Find the area of the rect
gayaneshka [121]
Hello,

It 's difficult to translate that in french.

l=16 ^(1/5)
w=2^(1/5)
Area=32^(1/5)=(2^5)^(1/5)=2 (in²)


7 0
3 years ago
Read 2 more answers
Suppose you can work at most a total of 25 hours per week. Baby-sitting, x, pays $6 per hour and working at the grocery store, y
Andrei [34K]

Answer:

<u><em>y=7  </em></u><u>number of hours at grocery store</u>

<u><em>x=18 </em></u><u>number of hours at baby- sitting</u>

Step-by-step explanation:

According to the information provided.

x is number of hours at baby- sitting

y is number of hours at grocery store

total number of hours worked

<em>1) </em><em>x+y =25</em>

total earn in a week

<em>2) </em><em>x*$6 + y* $9 = $171</em>

<em />

<em>from equation 1</em>

x+y=25

x= 25-y

<em>we place the above derived equation in equation 2 </em>

x*$6 + y* $9 = $171

(25-y)*$6 + y* $9 = $171

(25*6) -6y +9y =171

150+3y=171

3y=171-150

3y=21

<u><em>y=7  </em></u><u>number of hours at grocery store</u>

x= 25-y

x= 25-7

<u><em>x=18 </em></u><u>number of hours at baby- sitting</u>

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Answer:

Step-by-step explanation: subtract 160 and 125 and you’ll get 35

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I'm assuming you're supposed to find the chance of them drawing red or blue/
So the whole is 20, and red and blue combined is 14.So its 14/20, which is 70%, and as a simplified fraction it is 7/10.
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