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2 years ago

Find the fifth roots of 32(cos 280° + i sin 280°).

1 answer:

3 0

<span>Use the formula: r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ] where k = 0,1,2,3,4

</span><span>First 5th root: k = 0
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*0)/5 ) + i*sin( (280+360*0)/5 ) ]
2*[ cos( (280+360*0)/5 ) + i*sin( (280+360*0)/5 ) ]
2*[ cos( (280+0)/5 ) + i*sin( (280+0)/5 ) ]
2*[ cos( 280/5 ) + i*sin( 280/5 ) ]
2*[ cos( 56 ) + i*sin( 56 ) ]
-------------------------------------------------------------------
Second 5th root: k = 1
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*1)/5 ) + i*sin( (280+360*1)/5 ) ]
2*[ cos( (280+360*1)/5 ) + i*sin( (280+360*1)/5 ) ]
2*[ cos( (280+360)/5 ) + i*sin( (280+360)/5 ) ]
2*[ cos( 640/5 ) + i*sin( 640/5 ) ]
2*[ cos( 128 ) + i*sin( 128 ) ]
-------------------------------------------------------------------
Third 5th root: k = 2
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*2)/5 ) + i*sin( (280+360*2)/5 ) ]
2*[ cos( (280+360*2)/5 ) + i*sin( (280+360*2)/5 ) ]
2*[ cos( (280+720)/5 ) + i*sin( (280+720)/5 ) ]
2*[ cos( 1000/5 ) + i*sin( 1000/5 ) ]
2*[ cos( 200 ) + i*sin( 200 ) ]
-------------------------------------------------------------------
Fourth 5th root: k = 3
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*3)/5 ) + i*sin( (280+360*3)/5 ) ]
2*[ cos( (280+360*3)/5 ) + i*sin( (280+360*3)/5 ) ]
2*[ cos( (280+1080)/5 ) + i*sin( (280+1080)/5 ) ]
2*[ cos( 1360/5 ) + i*sin( 1360/5 ) ]
2*[ cos( 272 ) + i*sin( 272 ) ]
-------------------------------------------------------------------
Fifth 5th root: k = 4
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*4)/5 ) + i*sin( (280+360*4)/5 ) ]
2*[ cos( (280+360*4)/5 ) + i*sin( (280+360*4)/5 ) ]
2*[ cos( (280+1440)/5 ) + i*sin( (280+1440)/5 ) ]
2*[ cos( 1720/5 ) + i*sin( 1720/5 ) ]
2*[ cos( 344 ) + i*sin( 344 ) ]</span>

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2 years ago

The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes"† proposed using the Poisson distribution

oksian1 [2.3K]

Answer:

a. P(X ≤ 5) = 0.999

b. P(X > λ+λ) = P(X > 2) = 0.080

Step-by-step explanation:

We model this randome variable with a Poisson distribution, with parameter λ=1.

We have to calculate, using this distribution, P(X ≤ 5).

The probability of k pipeline failures can be calculated with the following equation:

$P(k)=\lambda^{k} \cdot e^{-\lambda}/k!=1^{k} \cdot e^{-1}/k!=e^{-1}/k!$

Then, we can calculate P(X ≤ 5) as:

$P(X\leq5)=P(0)+P(1)+P(2)+P(4)+P(5)\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\P(3)=1^{3} \cdot e^{-1}/3!=1*0.3679/6=0.061\\\\P(4)=1^{4} \cdot e^{-1}/4!=1*0.3679/24=0.015\\\\P(5)=1^{5} \cdot e^{-1}/5!=1*0.3679/120=0.003\\\\\\P(X\leq5)=0.368+0.368+0.184+0.061+0.015+0.003=0.999$

The standard deviation of the Poisson deistribution is equal to its parameter λ=1, so the probability that X exceeds its mean value by more than one standard deviation (X>1+1=2) can be calculated as:

$P(X>2)=1-(P(0)+P(1)+P(2))\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\\\P(X>2)=1-(0.368+0.368+0.184)=1-0.920=0.080$

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2 years ago

Every day a student randomly chooses a sandwich for lunch from a pile of wrapped sandwiches. If there are six kinds of sandwiche

balu736 [363]

Answer:

279,936 ways

Step-by-step explanation:

Every day the student has to chose a sandwich from the pile of 6 sandwiches. So this means the student has to make a choice from the 6 sandwiches for the 7 days. Since the order matters, this is a problem of permutations.

Daily the student has the option to chose from 6 sandwiches. So this means, for 7 days, he has to make a choice out of 6 options. Or in other words we can say, the student has to make selection from 6 objects 7 times.

So, the total number of ways to chose the sandwiches will be 6 x 6 x 6 x 6 x 6 x 6 x 6 = $6^{7}$

Alternate Method:

Since the repetition can occur in this case, i.e. a sandwich chosen on one day can also be chosen on other day, the following formula of permutations ca be used:

Number of ways = $n^{r}$

where n is the total number of choices available which is 6 in this case and r is the number of times the selection is to be made which 7 in this case. So,

The number of ways to chose a sandwich will be = $6^{7} = 279936$ ways

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1 year ago

Quadrilateral A’B’C’D’ is a dilation of quadrilateral ABCD about point P. Quadrilateral ABCD is shown. Side AB is labeled 5. Sid

Stolb23 [73]

Answer: The correct option is (A) reduction.

Step-by-step explanation: Given that the quadrilateral A'B'C'D' is a dilation of the quadrilateral ABCD.

As shown in the given figure, the lengths of the sides of quadrilateral ABCD are as follows:

AB = 5 units, BC = 4 units, CD = 10 units and DA = 6 units.

And, the lengths of the sides of quadrilateral A'B'C'D' are as follows:

$A'B'=1\dfrac{1}{4}=\dfrac{5}{4}~\textup{units},~~B'C'=1~\textup{units},~~C'D'=2\dfrac{1}{2}=\dfrac{5}{2}~\textup{units},\\\\D'A'=1\dfrac{1}{2}=\dfrac{3}{2}~\textup{units}.$

We know that the dilation will be an enlargement if the scale factor is greater than 1 and it will be a reduction if the scale factor is less than 1.

Now, the scale factor is given by

$S=\dfrac{\textup{length of a side of the dilated figure}}{\textup{length of the corresponding side of the original figure}}\\\\\\\Rightarrow S=\dfrac{A'B'}{AB}=\dfrac{\frac{5}{4}}{5}=\dfrac{5}{4\times5}=\dfrac{1}{4}$

Since the scale factor is less than 1, so the dilation will be a reduction.

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1 year ago

Read 2 more answers

What is the graph of the system y = -2x + 3 and 2x + + 4y = 8?

Maslowich

Answer:what are you solving it by

Step-by-step explanation:

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2 years ago