Connections between tan theta and tan(90 + theta)

1 answer:

7 0

Tan(90+ theta) = (tan90 + tantheta)/(1-tan90tantheta)

= ( sin90costheta + sintheta cos 90) / (cos90costheta - sin90sintheta))
= sin(90 + theta)/ cos(90+ theta) = costheta/-sintheta = - 1/tantheta

tan(90 + theta ) = -1/tantheta

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Please provide the answer?

Brums [2.3K]

Using the given table:

a) the average rate of change is 32.5 jobs/year.

b) the average rate of change is 12.5 jobs/year.

<h3>How to find the average rate of change?</h3>

For a function f(x), the average rate of change on an interval [a, b] is:

$\frac{f(b) - f(a)}{b - a}$

a) The average rate of change between 1997 and 1999 is:

$A = \frac{695 - 630}{1999 - 1997} = 32.5$

So the average rate of change is 32.5 jobs/year.

b) Now the interval is 1999 to 2001.

The rate this time is:

$A ' = \frac{720 - 695}{2001 - 1999} = 12.5$

So the average rate of change is 12.5 jobs/year.

If you want to learn more about average rates of change:

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The population, P(t), of China, in billions, can be approximated by1 P(t)=1.394(1.006)t, where t is the number of years since th

vitfil [10]

Answer:

At the start of 2014, the population was growing at 8.34 million people per year.

At the start of 2015, the population was growing at 8.39 million people per year.

Step-by-step explanation:

To find how fast was the population growing at the start of 2014 and at the start of 2015 we need to take the derivative of the function with respect to t.

The derivative shows by how much the function (the population, in this case) is changing when the variable you're deriving with respect to (time) increases one unit (one year).

We know that the population, P(t), of China, in billions, can be approximated by $P(t)=1.394(1.006)^t$

To find the derivative you need to:

$\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=\\\\\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\1.394\frac{d}{dt}\left(1.006^t\right)\\\\\mathrm{Apply\:the\:derivative\:exponent\:rule}:\quad \frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)\\\\1.394\cdot \:1.006^t\ln \left(1.006\right)\\\\\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=(1.394\cdot \ln \left(1.006\right))\cdot 1.006^t$

To find the population growing at the start of 2014 we say t = 0

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(0)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^0\\P(0)' = 0.00833901 \:Billion/year$

To find the population growing at the start of 2015 we say t = 1

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(1)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^1\\P(1)' = 0.00838904 \:Billion/year$