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dybincka [34]
3 years ago
10

Which of the following represents the zeros of f(x) = x3 − 12x2 + 47x − 60?

Mathematics
2 answers:
topjm [15]3 years ago
6 0

Answer:

x=3, 4, or 5

Step-by-step explanation:

x^3 − 12x^2 + 47x − 60=0

First we have to determine one value of x by hit and trial method for which , the above polynomial becomes 0

on trying random values of x , we sees that for x = 3 , the polynomial becomes 0

Hence (x-3) is one of the factor

Hence now let us move with the factorization part

The above polynomial can be re written as

x^2(x-3)-9x(x-3)+20(x-3)

(x-3)(x^2-9x+20)

(x-3)(x^2-5x-4x+20)

(x-3)(x(x-5)-4(x-5))

(x-3)(x-5)(x-4)

hence we have our f(x) = (x-3)(x-5)(x-4)

In order to find the zeroes we put f(x) =0

Hence

[tex](x-3)(x-5)(x-4)=0

Therefore

if (x-3)=0 , x=3

if (x-5) , x= 5

if (x-4), x=4

Nutka1998 [239]3 years ago
4 0
This actually should be reported. There are no choices. However there are ways of finding out what the zeros are and I will do that first to begin with. 

The first is a bit of a cheat. My calculator will do it, so I'm just going to get the answer from it. 

x can equal 3
x can be 5
x can be 4

So f(x)= a (x - 3)(x-5 )(x -4) is the basic equation. You need another point other than the zeros to determine the value of a.

Method 2
Get the graph. 
I've done that below. Basically it agrees with what my calculator says.

Method 3
You could factor it, but simple factoring methods don't work with this question.

Method 4
You could do this in a method similar to a quadratic, but I can't do that. It is on my to do list to lean though.

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Michael took the return trip at a velocity 33.75 miles per hour.

<h3>How fast did Michael drive in his return trip?</h3>

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3 0
2 years ago
Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

5 0
3 years ago
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