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hoa [83]
3 years ago
11

What is the leading coefficient of a third degree function that has an output of 221 when x=2, and has zeros of −15, 3i, and −3i

?
Mathematics
1 answer:
Vinil7 [7]3 years ago
3 0

Answer:

  1

Step-by-step explanation:

The function with the given zeros will factor as ...

  f(x) = a(x +15)(x^2 +9) . . . . with leading  coefficient 'a'

You have ...

  f(2) = 221 = a(2+15)(2^2+9) = a(17)(13) = 221a

Then a = 221/221 = 1

The leading coefficient is 1.

_____

<em>Additional comment</em>

As you know, a function with zero x=p has a factor of (x -p). The given zeros mean the function has factors (x -(-15)), (x -3i). and (x -(-3i)). The product of the last two factors is the difference of squares: (x^2 -(3i)^2) = (x^2 -(-9)) = (x^2 +9). This is how we arrived at the factorization shown above.

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Phoebe, Andy and Polly share £270.
Stella [2.4K]
*Given
Money of Phoebe            - 3 times as much as Andy
Money of Andy                - 2 times as much as Polly
Total money of Phoebe,  - <span>£270
</span>    Andy and Polly

*Solution

Let
B - Phoebe's money
A - Andy's money
L - Polly's money

1. The money of the Phoebe, Andy, and Polly, when added together would total <span>£270. Thus, 
</span>
                  B + A + L  = <span>£270                     (EQUATION 1)

2. Phoebe has three times as much money as Andy and this is expressed as
  
                  B = 3A

3. Andy has twice as much money as Polly and this is expressed as

                  A = 2L</span>                           (EQUATION 2)
<span>
4. This means that Phoebe has ____ as much money as Polly, 

                 B = 3A
                 B = 3 x (2L)
                 B = 6L                            </span>(EQUATION 3)<span>

This step allows us to eliminate the variables B and A in EQUATION 1 by expressing the equation in terms of Polly's money only. 

5. Substituting B with 6L, and A with 2L, EQUATION 1 becomes, 

                  6L + 2L + L = </span><span>£270
</span>                                9L = <span>£270
</span>                                  L = <span>£30

So, Polly has </span><span>£30. 

6. Substituting L into EQUATIONS 2 and 3 would give us the values for Andy's money and Phoebe's money, respectively. 
</span>
                  A = 2L                           
                  A = 2(£30)
                  A = £60

Andy has £60

                  B = 6L                         
                  B = 6(£30)
                  B = £180

Phoebe has £180

Therefore, Polly's money is £30, Andy's is £60, and Phoebe's is £180.
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