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4 years ago

Find the sum (3/7m-3+4n)+(2/7m-2m+6)

Mathematics

1 answer:

aleksley [76]4 years ago

4 0

Answer:

(5/7)m + 2n + 3

Step-by-step explanation:

Rewrite 3/7m as (3/7)m for added clarity, also 2/7m as (2/7)m.

Adding (3/7)m and (2/7)m results in (5/7)m.

Next, combine -3 and 6, which yields 3.

I expected that there'd be only one term in m in both quantities. (3/7)m - 3 + 4n does have only one term, but (2/7)m - 2m + 6) seems to have two m terms. I will assume that you actually meant (2/7)m - 2n + 6).

Combining like terms, we get (5/7)m + 3 + 4n - 2n, or

(5/7)m + 2n + 3

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max2010maxim [7]

Answer:

The complete question is:

At a university, 13% of students smoke.

a) Calculate the expected number of smokers in a random sample of 100 students from this university:

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Part a is easy, because is a random sample, we can expect that just 13% of these 100 students to be smokers, and 13% of 100 is 13, so we can expect 13 of those 100 students to be smokers.

b) This time we do not have a random sample, our sample is a sample of 15 students who go to the gym in the early morning, so our sample is biased. (And we do not know if this bias is related to smoking or not, and how that relationship is), so we can't use the same approach that we used in the previous part.

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3 years ago

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3 years ago

a. If cosθ=−45 where θ is in quadrant 3, find sin2θ. b. If cosθ=2√2 where θ is in quadrant 1, find cos2θ. c. If sinθ=817 where θ

sesenic [268]

Answer:

Part A) $sin(2\theta)=\frac{24}{25}$

Part B) $cos(2\theta)=0$

Part C) $tan(2\theta)=-\frac{240}{161}$

Step-by-step explanation:

Part A) we have $cos(\theta)=-\frac{4}{5}$

θ is in quadrant 3 ----> the sine is negative

Find $sin(2\theta)$

we know that

$sin(2\theta)=2sin(\theta)cos(\theta)$

Remember that

$cos^{2} (\theta)+sin^{2} (\theta)=1$

substitute

$(-\frac{4}{5})^{2}+sin^{2} (\theta)=1$

$(\frac{16}{25})+sin^{2} (\theta)=1$

$sin^{2} (\theta)=1-\frac{16}{25}$

$sin^{2} (\theta)=\frac{9}{25}$

$sin(\theta)=-\frac{3}{5}$ ---> remember that the sine is negative (3 quadrant)

Find $sin(2\theta)$

we have

$cos(\theta)=-\frac{4}{5}$

$sin(\theta)=-\frac{3}{5}$

$sin(2\theta)=2sin(\theta)cos(\theta)$

substitute

$sin(2\theta)=2(-\frac{3}{5})(-\frac{4}{5})$

$sin(2\theta)=\frac{24}{25}$

Part B) we have $cos(\theta)=\frac{\sqrt{2}}{2}$

θ is in quadrant 1

Find $cos(2\theta)$

we know that

$cos(2\theta)=2cos^{2} (\theta)-1$

substitute

$cos(2\theta)=2(\frac{\sqrt{2}}{2} )^{2}-1$

$cos(2\theta)=0$

Part C) we have $sin(\theta)=\frac{8}{17}$

θ is in quadrant 2 ----> the cosine is negative

Find $tan(2\theta)$

we know that

$tan(2\theta)=\frac{2tan(\theta)}{1-tan^{2} (\theta)}$

Remember that

$cos^{2} (\theta)+sin^{2} (\theta)=1$

substitute

$cos^{2} (\theta)+(\frac{8}{17})^{2}=1$

$cos^{2} (\theta)=1-\frac{64}{289}$

$cos^{2} (\theta)=\frac{225}{289}$

$cos(\theta)=-\frac{15}{17}$

Find $tan(\theta)$

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

substitute

$tan(\theta)=\frac{(8/17)}{(-15/17)}$

$tan(\theta)=-\frac{8}{15}$

Find $tan(2\theta)$

$tan(2\theta)=\frac{2tan(\theta)}{1-tan^{2} (\theta)}$

substitute

$tan(2\theta)=\frac{2(-\frac{8}{15})}{1-(-\frac{8}{15})^{2}}$

$tan(2\theta)=\frac{(-\frac{16}{15})}{1-(\frac{64}{225})}$

$tan(2\theta)=\frac{(-\frac{16}{15})}{1-\frac{64}{225}}$

$tan(2\theta)=\frac{(-\frac{16}{15})}{\frac{161}{225}}$

$tan(2\theta)=-\frac{240}{161}$

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Answer:

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Step-by-step explanation:

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