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UNO [17]
3 years ago
15

Which zero pair could be added to the function f(x)=x^2+12x+6 so that the function can be written in vertex form?

Mathematics
1 answer:
andrew11 [14]3 years ago
4 0
<span>f(x) = x</span>² <span>+ 12x + 6  </span>→ y = x² + 12x + 6<span>

Let us convert the standard form into vertex form.

1) Complete the squares. Isolate x</span>² and x terms.
<span>y - 6 = x</span>² + 12x
<span>
2) Create the perfect square trinomial. Whatever number is added on one side must also be added on the other side. 

y - 6 + 36 = x</span>² + 12x + 36<span>
y + 30 = (x + 6)</span>²
<span>y = (x + 6)</span>² - 30  ← Vertex form
<span>
To check:

y = (x + 6) (x + 6) - 30
y = x</span>² + 6x + 6x + 36 - 30
<span>y = x</span>² + 12x + 6<span>

The zero that could be added to the given function is 36, -36</span>
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Solve for x. 0.6x = 1.2x + 6
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8 0
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Divide. Write the quotient in lowest terms. 3/5. Devide by 2 1/2 how much ????
dybincka [34]

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Step-by-step explanation:

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6 0
3 years ago
The nth term of a sequence is 3n^2- 1
Thepotemich [5.8K]

Answer: The number that belongs to both sequences is 26.

Step-by-step explanation:

We have two sequences, let's call one as A and the other as B.

The n-th term of sequence A is written as:

aₙ = 3*n^2 - 1

the nth term of sequence B is written as:

bₙ = 30 - n^2

We want to find a term that belongs to both sequences, (it can be for different integers, we can use n for sequence A and x for sequence B)

Then we want to find:

aₙ = bₓ

where n and x are integer numbers.

Then we will heave:

3*n^2 - 1 = 30 - x^2

To find the pair, we could isolate one of the variables, then input different integers in the other variable and see if the outcome is also an integer.

Let's isolate n.

3*n^2 = 30 - x^2 + 1

3*n^2 = 31 - x^2

n^2 = (31 - x^2)/3

n = √(  (31 - x^2)/3)

Now let's input different values for x, and see if the outcome is also an integer, notice that x is in a negative term inside a square root, then we have only a few values of x such that the equation can be true.

Then let's start with x = 1.

n(1) = √(  (31 - 1^2)/3) = √(30/3) = √10

We know that √10 is not an integer.

now with x = 2,

n(2) = √(  (31 - 2^2)/3)  = √( (31 - 4)/3) = √(27/3) = √9 = 3

then if x = 2, we have n = 3.

Both of them are integers, then we get:

a₂ = 3*(3)^2 - 1 = 27 - 1 = 26

b₃ = 30 - 2^2 = 30 - 4 = 26

The number that belongs to both sequences is 26.

5 0
3 years ago
(Multiple Choice Worth 4 points)
fgiga [73]
<u>Rectangle A</u>
P = 2l + 2w
P = 2(3x + 2) + 2(2x - 1)
P = 2(3x) + 2(2) + 2(2x) - 2(1)
P = 6x + 4 + 4x - 2
P = 6x + 4x + 4 - 2
P = 10x + 2
<u>
Rectangle B</u>
P = 2l + 2w
P = 2(x + 5) + 2(5x - 1)
P = 2(x) + 2(5) + 2(5x) - 2(1)
P = 2x + 10 + 10x - 2
P = 2x + 10x + 10 - 2
P = 12x + 8

<u>Rectangle B - Rectangle A</u>
(12x + 8) - (10x + 2)
12x - 10x + 8 - 2
2x + 6

The correct answer is B.
5 0
3 years ago
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