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Lady bird [3.3K]
4 years ago
11

The quadratic equation by completing the square x^2+6x=18

Mathematics
1 answer:
g100num [7]4 years ago
3 0

This technique is valid only when the coefficient of x2 is 1.

1)   Transpose the constant term to the right

x2 + 6x = −2.

<span> 2)  <span>Add a square number to both sides -- add the square of half the coefficient of x.  In this case, add the square of half of 6; that is, add the square of 3, which is 9:</span></span>

x2 + 6x + 9 = −2 + 9.

The left-hand side is now the perfect square of  (x + 3).

(x + 3)2  =  7.

3 is half of the coefficient 6.

That equation has the form

<span><span><span>a2</span> = b</span>  which implies<span>a = <span>±.</span></span>         Therefore,<span><span>x + 3</span> = ±</span> <span>x = <span>−3 ±.</span></span></span>

That is, the solutions to

x2 + 6x + 2  =  0


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4x + 3y = 16<br> 7x+6y=31<br> O A. (1,3)<br> B. (1.4)<br> C. (4,0)<br> OD. (4,3)<br> SUBMIT
klasskru [66]

Answer:

B

Step-by-step explanation:

Given the 2 equations

4x + 3y = 16 → (1)

7x + 6y = 31 → (2)

Multiplying (1) by - 2 and adding to (2) will eliminate the term in y

- 8x - 6y = - 32 → (3)

Add (2) and (3) term by term

- x = - 1 ( multiply both sides by - 1 )

x = 1

Substitute x = 1 in either of the 2 equations and solve for y

Substituting in (1)

4(1) + 3y = 16

4 + 3y = 16 ( subtract 4 from both sides )

3y = 12 ( divide both sides by 3 )

y = 4

Solution is (1, 4 ) → B

4 0
3 years ago
Please answer these two questions
Volgvan
The correct answer is two or four
6 0
3 years ago
Plsss help<br> solve for x
Gnoma [55]
8/7= 7/2x
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5 0
3 years ago
Is Event B dependent or independent of Event A? A: A green ball is drawn from a box with five balls and placed next to the box.
Zarrin [17]

Answer:

Step-by-step explanation:

Given that events are as follows:

A: A green ball is drawn from a box with five balls and placed next to the box. B: A red ball is drawn next and placed next to the green one

i.e. drawings of balls are made without replacement

Suppose there are m green balls and n red balls, where m+n =5

Prob of drawing green ball =m/5

After I draw made remaining are n red balls and m-1 green balls

Prob of drawing red ball now =n/4

Prob for drawing one green and one red simultaneously = \frac{mC1*nC1}{5C2} \\= \frac{mn}{10}

Product of the two probabilities = \frac{m}{5} *\frac{n}{4} \\=\frac{mn}{20}

6 0
4 years ago
What is the best estimate for the sum of 3/8and1/12
hichkok12 [17]
First all you have to do is add 3/8 and 1/12 which is 4/20 .

Now when you estimate 4/20, it will be 1/5
5 0
4 years ago
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