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Vladimir79 [104]
3 years ago
9

Which two values of x are roots of the polynomial below 3x²-3x+1​

Mathematics
2 answers:
pshichka [43]3 years ago
6 0

Answer:

x=\frac{1}{2} +\sqrt{\frac{1}{9} } i , \frac{1}{2} -\sqrt{\frac{1}{9} } i

Step-by-step explanation:

Just another way to write the answer on top of me.

ArbitrLikvidat [17]3 years ago
5 0

Answer:  x  = (3 − i√3)/6 and (3 + i√3)/6

Step-by-step explanation:3x² − 3x + 1 = 0

∴ 3x² − 3x = -1

∴ x² − x = -1/3

∴ x² − x + (-1/2)² = (-1/2)² − 1/3

given x² + bx + (b/2)² = (x + b/2)²

∴ (x − 1/2)² = 1/4 − 1/3

∴ (x − 1/2)² = 3/12 − 4/12

∴ (x − 1/2)² = -1/12

∴ x − 1/2 = ±√(-1/12)

This tells us there are no real roots and if you need real number solutions we stop here

∴ x − 1/2 = ±i/(2√3)

∴ x − 1/2 = ±i√(3)/6

∴ x  = 1/2 ± i√(3)/6

∴ x  = 3/6 ± i√(3)/6

∴ x  = (3 − i√3)/6 and (3 + i√3)/6 <= the 2 complex roots

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Find the reciprocals of the following numbers: 8, 2/7 , 1 1/5 , 0.3.
Phoenix [80]
1/8, 7/2, 5/6, 10/3

just swap numerator and denominator for each
8 0
3 years ago
How many 4-digit odd numbers can you make using the digits 1 to 7 if the numbers must be less than 6000? No digits are repeated.
bixtya [17]

Answer:

Step-by-step explanation:he first digit must be a 6, 7, or 8, because the number has to be bigger than 6,000.

The last digit must be a 1, 3, 5, or 7, because the number has to be odd.

So if the first digit is 6 or 8, there are 4 odd-number options for the last digit. If the first digit is a 7, then there are only 3 options for the last digit, as 7 can’t be used twice.

The other two digits can be anything that’s not already used. So for each first-and-last combination of digits mentioned above, there will be six options available for one of the middle digits, and then for each of those 3-digit combinations, five options will be available for the final remaining digit.

At this point, we have everything we need to express this as an equation:

x = (2 x 4 + 1 x 3) x 6 x 5

And then we can simplify and solve:

x = (8 + 3) x 30

x = 11 x 30

x = 330

6 0
2 years ago
An education researcher claims that 58​% of college students work​ year-round. In a random sample of 400 college​ students, 232
Hatshy [7]

Answer:

The proportion of college students who work​ year-round is 58%.

Step-by-step explanation:

The claim made by the education researcher is that 58​% of college students work​ year-round.

A random sample of 400 college​ students, 232 say they work​ year-round.

To test the researcher's claim use a one-proportion <em>z</em>-test.

The hypothesis can be defined as follows:

<em>H</em>₀: The proportion of college students who work​ year-round is 58%, i.e. <em>p</em> = 0.58.

<em>Hₐ</em>: The proportion of college students who work​ year-round is 58%, i.e. <em>p</em> ≠ 0.58. C

Compute the sample proportion as follows:

 \hat p=\frac{232}{400}=0.58

Compute the test statistic value as follows:

 z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.58-0.58}{\sqrt{\frac{0.58(1-0.58)}{400}}}=0

The test statistic value is 0.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

Compute the p-value for the two-tailed test as follows:

 p-value=2\times P(z

*Use a z-table for the probability.

The p-value of the test is 1.

The p-value of the test is very large when compared to the significance level.

The null hypothesis will not be rejected.

Thus, it can be concluded that the proportion of college students who work​ year-round is 58%.

6 0
3 years ago
Simplify: −3 • −26<br><br> A. −78<br> B. −68<br> C. 68<br> D. 78
hammer [34]
-3 x -26 = 78

meaning the answer is D.

The reason the answer is D, is because two negatives when multiplied or divided become positive. Only when multiplied or divided does this occur.
3 0
1 year ago
Help pls will mark BRAINLY
Daniel [21]

Answer:

place your dots on the grid

Step-by-step explanation:

for example r would beeee uhm 3 y x 2

6 0
2 years ago
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