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2 years ago

Simplify sin q + cosq cotq.

1 answer:

3 0

$\sin q+\cos q\cot q=\sin q+\cos q\cdot\dfrac{\cos q}{\sin q}=\sin q+\dfrac{\cos^2q}{\sin q}\\\\=\dfrac{sin^2q}{\sin q}+\dfrac{\cos^2q}{\sin q}=\dfrac{\sin^2q+\cos^2q}{\sin q}=\dfrac{1}{\sin q}=\csc q\\\\Used:\\\cot q=\dfrac{\cos q}{\sin q}\\\\\sin^2q+\cos^2q=1\\\\\csc q=\dfrac{1}{\sin q}$

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A workers regular hourly rate of pay is $13. What is the workers total pay for working 40 regulars and 3 hours overtime at a rat

ANTONII [103]

Answer:

578.50

Step-by-step explanation:

The worker worked 40 hours at 13 dollars

40 * 13 =520

Then they worked 3 hours at 13(1.5) or 19.50

3*19.50 =58.5

The total amount of money earned is regular time plus overtime

520 + 58.5 =578.50

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2 years ago

The following data represent the pH of rain for a random sample of 12 rain dates. a normal probability plot suggest the data cou

babunello [35]

Answer:

Step-by-step explanation:

a) Estimate for population mean = Sum/n = $\frac{59.1}{12} =4.925$

b) Variance = 0.099208

Std dev = 0.314974

Std error = 0.0950

For 95% margin of error = 1.96*Std error

=0.1861

Confidence interval = $(4.925-0.1861, 4.925+0.1861)\\=(4.7389, 5.1111)$

Interpretation of confidence interval:

A) if repeated samoles are taken, 95% of them will have a sample pH of rain water between [ ] & [ ].

For 99% CI, z value = 2.59

Conf interval = (4.6790, 5.1710)

C)if repeated samoles are taken, 99% of them will have a sample pH of rain water between [ ] & [ ].

As the level of confidence increases l, the width of the interval[ increases] this makes sense since the [ margin of error] [increases]

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2 years ago

What is the slope of the line that passes through the pair of points? (1, 7), (10, 1)

Harrizon [31]

I believe it would be C<span />

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2 years ago

For the function below, find a formula for the upper sum obtained by dividing the interval [a comma b ][a,b] into n equal subin

Vlad [161]

Answer:

See below

Step-by-step explanation:

We start by dividing the interval [0,4] into n sub-intervals of length 4/n

$[0,\displaystyle\frac{4}{n}],[\displaystyle\frac{4}{n},\displaystyle\frac{2*4}{n}],[\displaystyle\frac{2*4}{n},\displaystyle\frac{3*4}{n}],...,[\displaystyle\frac{(n-1)*4}{n},4]$

Since f is increasing in the interval [0,4], the upper sum is obtained by evaluating f at the right end of each sub-interval multiplied by 4/n.

Geometrically, these are the areas of the rectangles whose height is f evaluated at the right end of the interval and base 4/n (see picture)

$\displaystyle\frac{4}{n}f(\displaystyle\frac{1*4}{n})+\displaystyle\frac{4}{n}f(\displaystyle\frac{2*4}{n})+...+\displaystyle\frac{4}{n}f(\displaystyle\frac{n*4}{n})=\\\\=\displaystyle\frac{4}{n}((\displaystyle\frac{1*4}{n})^2+3+(\displaystyle\frac{2*4}{n})^2+3+...+(\displaystyle\frac{n*4}{n})^2+3)=\\\\\displaystyle\frac{4}{n}((1^2+2^2+...+n^2)\displaystyle\frac{4^2}{n^2}+3n)=\\\\\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12$

but

$1^2+2^2+...+n^2=\displaystyle\frac{n(n+1)(2n+1)}{6}$

so the upper sum equals

$\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12=\displaystyle\frac{4^3}{n^3}\displaystyle\frac{n(n+1)(2n+1)}{6}+12=\\\\\displaystyle\frac{4^3}{6}(2+\displaystyle\frac{3}{n}+\displaystyle\frac{1}{n^2})+12$