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Oduvanchick [21]
2 years ago
6

How do you simplify the following: (sqrt(6) + sqrt(2))/2

Mathematics
1 answer:
nignag [31]2 years ago
8 0
Not all expressions of this form can be simplified; but if one can, we would like the square root to have the form sqrt(a) + sqrt(b) or sqrt(a) - sqrt(b) where a and b are positive rational numbers, and a > b to give the positive square root. If we assume that sqrt(2-sqrt(3)) has this form, then this must be true: (sqrt(a) +- sqrt(b))^2 = 2 - sqrt(3) Expanding, we get a + b +- 2 sqrt(ab) = 2 - sqrt(3) Equating the rational and irrational parts of this equation, a + b = 2 +-2 sqrt(ab) = -sqrt(3) The second equation here tells us that we will have to choose the negative sign in order to have a solution, since square roots have to be positive. If we now square this equation, we have a + b = 2 4ab = 3 Solving these equations by substituting in the second, we get 4a(2 - a) = 3 4a^2 - 8a - 3 = 0 8 +- sqrt(64 + 48) a = ------------------ = 1/2 or 3/2 8 and b = 2 - a = 3/2 or 1/2. One choice here will give the positive root, the other the negative root. We take a > b, so that a = 3/2 and b = 1/2. Consequently, our solution is sqrt(2 - sqrt(3)) = sqrt(3/2) - sqrt(1/2) = (sqrt(6) - sqrt(2))/2 Looking for information on this technique, which I didn't recall being taught, I found this page, which together with some rather advanced techniques includes a quote from an old algebra text about the method:
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Serggg [28]

Answer:

1280 sq inches

Step-by-step explanation:

A = LW

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A = 1280 sq inches

7 0
3 years ago
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A sample of 50 drills had a mean lifetime of 12.17 holes drilled when drilling a low-carbon steel. Assume the population standar
zmey [24]

Using the z-distribution, it is found that the 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).

We are given the <u>standard deviation for the population</u>, hence, the z-distribution is used. The parameters for the interval is:

  • Sample mean of \overline{x} = 12.17
  • Population standard deviation of \sigma = 6.37
  • Sample size of n = 50.

The margin of error is:

M = z\frac{\sigma}{\sqrt{n}}

In which z is the critical value.

We have to find the critical value, which is z with a p-value of \frac{1 + \alpha}{2}, in which \alpha is the confidence level.

In this problem, \alpha = 0.95, thus, z with a p-value of \frac{1 + 0.95}{2} = 0.975, which means that it is z = 1.96.

Then:

M = 1.96\frac{6.37}{\sqrt{50}} = 1.77

The confidence interval is the <u>sample mean plus/minutes the margin of error</u>, hence:

\overline{x} - M = 12.17 - 1.77 = 10.4

\overline{x} + M = 12.17 + 1.77 = 13.94

The 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).

A similar problem is given at brainly.com/question/22596713

6 0
3 years ago
56 + 32 = 8(7 + *blank*)<br><br> Enter the unknown number that makes the equation true.
kati45 [8]
The correct answer is 4
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3 years ago
Oml please don’t say anything that doesn’t help,
Veseljchak [2.6K]

Answer:

first convert mixed fraction to improper fraction then solve by taking LCM

hope the above process helps

6 0
2 years ago
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SOLVE FOR X. pls i need help
Lyrx [107]
X=8
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Turns into: 70x -10 = 66x+22. Then just solve for x
5 0
3 years ago
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