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3 years ago

Using basic trigonometric identities simplify the

Mathematics

1 answer:

katrin2010 [14]3 years ago

6 0

Answer:

Step-by-step explanation:

Using the trigonometric identities

cot x = $\frac{cosx}{sinx}$ and tan x = $\frac{sinx}{cosx}$ thus

$\frac{cos0}{cot0}$ + tanΘ × cosΘ

= $\frac{cos0}{\frac{cos0}{sin0} }$ + $\frac{sin0}{cos0}$ × cosΘ

= cosΘ × $\frac{sin0}{cos0}$ + sinΘ

= sinΘ + sinΘ

= 2sinΘ → b

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A quantity with an initial value of 9100 grows continuously at a rate of 0.85% per hour. What is the value of the quantity after

ycow [4]

Answer:

$10,962,54$

Step-by-step explanation:

we know that

The equation of a exponential growth is equal to

$y=a(1+r)^x$

where

a is the initial value

r is the rate of growth

x is the time in hours

y is the value of the quantity

we have

$a=9,100\\r=0.85\%=0.85/100=0.0085$

substitute

$y=9,100(1+0.0085)^x$

$y=9,100(1..0085)^x$

For x=22 hours

substitute

$y=9,100(1..0085)^{22} =10,962,54$

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3 years ago

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Tanya [424]

Answer:

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Step-by-step explanation:

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8 0

3 years ago

Read 2 more answers

Combine and simplify the third set of terms 2+15a-2

Finger [1]

<span>2 + 15a - 2
= 15a (because 2 - 2 = 0)

hope it helps</span>

6 0

3 years ago

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Otrada [13]

Answer:

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Step-by-step explanation:

4 0

2 years ago

A population has a standard deviation of 5.5. What is the standard error of the sampling distribution if the sample size is 81?

VladimirAG [237]

Answer:

$\sigma = 5.5$

And we have a sample size of n =81. We want to estimate the standard error of the sampling distribution $\bar X$ and for this case we know that the distribution is given by:

$\bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})$

And the standard error would be:

$\sigma_{\bar x}= \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$\sigma_{\bar x}=\frac{5.5}{\sqrt{81}}= 0.611$

Step-by-step explanation:

For this case we know the population deviation given by:

$\sigma = 5.5$

And we have a sample size of n =81. We want to estimate the standard error of the sampling distribution $\bar X$ and for this case we know that the distribution is given by:

$\bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})$

And the standard error would be:

$\sigma_{\bar x}= \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$\sigma_{\bar x}=\frac{5.5}{\sqrt{81}}= 0.611$

4 0

3 years ago