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Elenna [48]
4 years ago
9

Write the standard form of the simplest polynomial with the roots 2, 3, and -3

Mathematics
1 answer:
koban [17]4 years ago
5 0

\text{If}\ x_1,\ x_2\ \text{and}\ x_3\ \text{are roots of a polynomial, then we can write that polynomial}\\\text{in form}\ (x-x_1)(x-x_2)(x-x_3).\\\\\text{We have\ the\ roots:}\ 2,\ 3\ \text{and}\ -3.\\\\w(x)=(x-2)(x-3)(x-(-3))=(x-2)\underbrace{(x-3)(x+3)}_{use\ a^2-b^2=(a-b)(a+b)}\\\\=(x-2)(x^2-9^2)=(x-2)(x^2-9)=(x)(x^2)+(x)(-9)+(-2)(x^2)+(-2)(-9)\\\\=x^3-9x-2x^2+18=\boxed{x^3-2x^2-9x+18}

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Prove that the value of expression 9^7+3^12 is divisible by 90.
snow_tiger [21]
9^7=4782969 and 3^12=531441 and when you add 4782969 and 531441 together you get 5314410. Then divide that number by 90 and get 59049
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Help plz I dont no 500+500 =(​
tino4ka555 [31]

Answer:

500+500=1,000

Step-by-step explanation:

I hope this helps!

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Thepotemich [5.8K]
2/3 times what=6 and 2/3
convert to improper
6 and 2/3=6+2/3=18/3+2/3=20/3
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4 years ago
The following measurements (in picocuries per liter) were recorded by a set of argon gas detectors installed in a research facil
Reptile [31]

Answer:

The 95% confidence interval for the mean level of argon gas present in the facility is (374.4, 401.7).

Step-by-step explanation:

Before building the confidence interval, we have to find the sample mean and the sample standard deviation.

Sample mean:

\overline{x} = \frac{381.3+394.8+396.1+380}{4} = 388.05

Sample standard deviation:

s = \sqrt{\frac{(381.3-388.05)^2+(394.8-388.05)^2+(396.1-388.05)^2+(380-388.05)^2}{3}} = 8.58

Confidence interval:

We have the standard deviation for the sample, so the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 4 - 1 = 3

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 3 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 3.1824

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 3.1824\frac{8.58}{\sqrt{4}} = 13.65

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 388.05 - 13.65 = 374.4

The upper end of the interval is the sample mean added to M. So it is 388.05 + 13.65 = 401.7

The 95% confidence interval for the mean level of argon gas present in the facility is (374.4, 401.7).

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3 years ago
The private school enrollment of a city is 12.000. Enrollment is increasing at a rate of 4.5% annually. What would the
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Answer:

28940.568

Step-by-step explanation:

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