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adelina 88 [10]
3 years ago
5

0/1 Solve the equation by completing the square. Round your answer to the nearest hundredth, if necessary x^2+18x=7

Mathematics
1 answer:
11111nata11111 [884]3 years ago
7 0

Answer:

The solution to the equation is x = 0.38 or -18.38

Step-by-step explanation:

x^2+18x=7 can be properly written as

x^{2} +18x =7

To solve by completing the square, divide the coefficient of x by 2, square and then add to both sides

i.e 18/2 = 9

Then add 9² to both sides of the equation

x^{2} +18x + 9^{2} = 7 + 9^{2}

This becomes

(x+9)^{2} = 7 + 81

(x+9)^{2} = 88

x+9 = \pm \sqrt{88}

x = -9 \pm\sqrt{88}

x = -9 \pm 9.38

∴ x = -9+9.38 or -9-9.38

x = 0.38 or -18.38

Hence, x = 0.38 or -18.38

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Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%. Find
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Answer:

1) \text{P(at least one boy and one girl)}=\frac{3}{4}

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3) \text{P(at least two girls)}=\frac{1}{2}

Step-by-step explanation:

Given : Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%.

To  Find : The probability of each event.  

1) P(at least one boy and one girl)

2) P(two boys and one girl)

3) P(at least two girls)        

Solution :

Let's represent a boy with B and a girl with G

Mr. and Mrs. Romero are expecting triplets.

The possibility of having triplet is

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Total outcome = 8

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

1) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, BGG, GBB, GBG, GGB=6

\text{P(at least one boy and one girl)}=\frac{6}{8}

\text{P(at least one boy and one girl)}=\frac{3}{4}

2) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, GBB=3

\text{P(at least one boy and one girl)}=\frac{3}{8}

3) P(at least two girls)

Favorable outcome = BGG, GBG, GGB, GGG=4

\text{P(at least two girls)}=\frac{4}{8}

\text{P(at least two girls)}=\frac{1}{2}

4 0
3 years ago
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