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saveliy_v [14]
2 years ago
9

Solve the quadratic equation by completing the square. x²+18x+79=0

Mathematics
1 answer:
pogonyaev2 years ago
8 0

x^2 + 18x=-79 \\ \\ x^2 + 18x+81=2 \\ \\ \boxed{(x+9)^2=2} \\ \\ x+9=\pm \sqrt 2 \\ \\ x=-9 \pm \sqrt{2} \\ \\ \boxed{x=-9+\sqrt{2}, -9-\sqrt{2}}

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Answer:

8

Step-by-step explanation:

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8x=64

x=8

8 0
3 years ago
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Jim has 2 gems. He got 2 more on Monday and 5 more on Tuesday. How many does he have now?
Lunna [17]

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9 gems

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Part A:we know  p(t)=6t and also
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4 years ago
What is the exact value of sin(5π/12)? <br><br>(please give steps)​
snow_tiger [21]

Answer:

0.966053748

Step-by-step explanation:

\sin \bigg( \frac{5\pi}{12} \bigg) \\  =  \sin \bigg( \frac{5 \times 180 \degree}{12} \bigg) \\  =   \sin ( 5 \times 15 \degree) \\  =  \sin ( 75 \degree) \\  = \sin ( 45 \degree + 30 \degree) \\  = \sin  45 \degree  \times  \cos 30 \degree + \cos  45 \degree  \times  \sin 30 \degree \\  \\  =  \frac{1}{ \sqrt{2} }  \times   \frac{ \sqrt{3} }{2}  +  \frac{1}{ \sqrt{2} }  \times  \frac{1}{2}  \\  \\  =  \frac{ \sqrt{3} }{2 \sqrt{2} }  +  \frac{1}{2 \sqrt{2} }  \\  \\  =  \frac{ \sqrt{3}  + 1}{2 \sqrt{2} }  \\  \\  =  \frac{1.732 + 1}{2 \times 1.414}  \\  \\  =  \frac{2.732}{2.828}  \\  \\  = 0.966053748

7 0
3 years ago
A company that produces fine crystal knows from experience that 17% of its goblets have cosmetic flaws and must be classified as
Alex73 [517]

Answer:

0.3891 = 38.91% probability that only one is a second

Step-by-step explanation:

For each globet, there are only two possible outcoes. Either they have cosmetic flaws, or they do not. The probability of a goblet having a cosmetic flaw is independent of other globets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

17% of its goblets have cosmetic flaws and must be classified as "seconds."

This means that p = 0.17

Among seven randomly selected goblets, how likely is it that only one is a second

This is P(X = 1) when n = 7. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{7,1}.(0.17)^{1}.(0.83)^{6} = 0.3891

0.3891 = 38.91% probability that only one is a second

7 0
3 years ago
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