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3 years ago

What is exacly 5/6 of a full rotation

Mathematics

2 answers:

solmaris [256]3 years ago

7 0

Full rotation= 360 degrees

360(5/6)= 300 degrees

The answer is 300 degrees.

Hope this helps!

Katen [24]3 years ago

6 0

300 degrees is the answer

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Answer:

scalene triangle

Step-by-step explanation:

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3 years ago

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2 years ago

Simplify <br> Rewrite the expression in the form z^n<br><br> z^5•z^6= ______

gtnhenbr [62]

Answer:

z²^30

Step-by-step explanation:

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3 years ago

the performance score of 10 adults is recorded, and the results are 83 87 90 92 93 100 104 111 115 121 find the standard deviati

luda_lava [24]

Answer:

The standard deviation of the data set is $\sigma = 12.7906$ .

Step-by-step explanation:

The Standard Deviation is a measure of how spread out numbers are. Its symbol is σ (the greek letter sigma)

To find the standard deviation of the following data set

$\begin{array}{cccccccc}83&87&90&92&93&100&104&111\\115&121&&&&&&\end{array}$

we use the following formula

$\sigma = \sqrt{ \frac{ \sum{\left(x_i - \overline{X}\right)^2 }}{n-1} }$

Step 1: Find the mean $\left( \overline{X} \right)$ .

The mean of a data set is the sum of the terms divided by the total number of terms. Using math notation we have:

$Mean = \frac{Sum ~ of ~ terms}{Number ~ of ~ terms}$

$Mean = \frac{Sum ~ of ~ terms}{Number ~ of ~ terms}=\frac{83+87+90+92+93+100+104+111+115+121}{10} \\\\Mean = \frac{996}{10} =\frac{498}{5}=99.6$

Step 2: Create the below table.

Step 3: Find the sum of numbers in the last column to get.

$\sum{\left(x_i - \overline{X}\right)^2} = 1472.4$

Step 4: Calculate σ using the above formula.

$\sigma = \sqrt{ \frac{ \sum{\left(x_i - \overline{X}\right)^2 }}{n-1} } = \sqrt{ \frac{ 1472.4 }{ 10 - 1} } \approx 12.7906$

3 0

3 years ago

If α and β are the zeros of the quadratic polynomial f(x) = 3x2–4x + 5, find a polynomialwhose zeros are 2α + 3β and 3α + 2β.

Lelechka [254]

Answer:

$\boxed{\sf \ \ \ 3x^2-20x+37\ \ \ }$

Step-by-step explanation:

Hello,

a and b are the zeros, we can say that

$f(x)=3(x^2-\dfrac{4}{3}x+\dfrac{5}{3}) = 3(x-a)(x-b)=3(x-(a+b)x+ab)$

So we can say that

$a+b=\dfrac{4}{3}\\ab=\dfrac{5}{3}$

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b

for instance we can write

$(x-2a-3b)(x-3a-2b)=x^2-(2a+3b+3a+2b)x+(2a+3b)(3a+2b)\\= x^2-5(a+b)x+6a^2+6b^2+9ab+4ab$

and we can notice that

$a^2+b^2=(a+b)^2-2ab$ so

$(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab$

it comes

$x^2-5*\dfrac{4}{3}x+6(\dfrac{4}{3})^2+\dfrac{5}{3}$

multiply by 3

$3x^2-20x+2*16+5=3x^2-20x+37$

4 0

3 years ago