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OverLord2011 [107]
3 years ago
12

If a polynomial function f(x) has roots –8, 1, and 6i, what must also be a root of f(x)? A. –6

Mathematics
2 answers:
-BARSIC- [3]3 years ago
7 0

Answer:

B. –6i

Step-by-step explanation:

Complex roots come in pairs.  The pairs are complex conjugates.  If you have a+bi, then you will have a-bi

Since we have 6i,  the other root must be -6i

anyanavicka [17]3 years ago
6 0

Answer:

-6i

Step-by-step explanation:

Complex roots always come in pairs, and those pairs are made up of a positive and a negative version. If 6i is a root, then its negative value, -6i, is also a root.

If you want to know the reasoning, it's along these lines: to even get a complex/imaginary root, we take the square root of a negative value. When you take the square root of any value, your answer is always "plus or minus" whatever the value is. The same thing holds for complex roots. In this case, the polynomial function likely factored to f(x) = (x+8)(x-1)(x^2+36). To solve that equation, you set every factor equal to zero and solve for the x's.

x + 8 = 0

x = -8

x - 1 = 0

x = 1

x^2 + 36 = 0

x^2 = -36 ... take the square root of both sides to get x alone

x = √-36 ... square root of an imaginary number produces the usual square root and an "i"

x = ±6i

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Answer: I believe the answer is 0.1875. Must be a typo here, because...

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After parentheses are taken out. 3x3%32= No exponents, so multiply or divide, whatever comes first.

6%32 is 0.1875

5 0
3 years ago
Consider the following theorem and proof.Theorem: The number âš2 is not rational number.Proof: Let's suppose âš2 is a rational n
ololo11 [35]

Answer:

The statement "Both of the numbers a and b cannot be even." is justified by the fact that a/b is simplified lowest terms

Step-by-step explanation:

We need to  show that the √2 is an irrational number.

And from the given steps of proof stated in the question, we need to find the assumption that justifies the fact : " Both of the number a and b cannot be even".

First take the given options :

Option a : √2 is a rational number

√2 being an rational or irrational has no relation of a and b to be even or odd. So, this option is rejected.

Option B : a/b is simplified lowest terms

This shows that a and b are not even because if a and b are even then a/b can be simplified in other lowest term.

Option c : √2 is a irrational number

Similarly, By using the inverse part of Option A, option c is also rejected.

Option d : The fact that b divides a evenly

This only shows that the a is even. This does not give any idea about b is even or not. So option D is also rejected.

Option E : The fact that a and b are whole numbers

This fact does not imply that the a and b are even or odd. So option E is also rejected.

Hence, The statement "Both of the numbers a and b cannot be even." is justified by the fact that a/b is simplified lowest terms

7 0
3 years ago
find real numbers a, b, and c so that the graph of the function y=ax^2+c contains the points (-1,6), (2,7), and (0,1)
Paraphin [41]
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6 = a + c
7 = 4a + c
1 = c
 
<u>c = 1
</u>
<u /><u />
If c =1,

6 = a + 1
<u>a = 5
</u>
<u /><u />
This doesn't work in the second equation, so the quadratic that goes through these points is not in the form y = ax^2 + bx + c
Was there supposed to be a b in the equation?




5 0
3 years ago
Joe bought a box of laundry detergent that contains 195 scoops.
7nadin3 [17]

Answer:

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Step-by-step explanation:

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8 0
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