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3 years ago

Help please dghsshjhjs

Mathematics

1 answer:

ser-zykov [4K]3 years ago

6 0

Answer:

Step-by-step explanation:

There is more than one way to solve this; the easiest way is to plug in some numbers for x. We can see the roots are (3,0) and (7,0) and the lowest point is (5,0). So when we plug in 3 or 7, we should get zero.

B: (3-3)(3-7)=0

D: 3(3-3)(3-7)=0

So we have two possible solutions, So to check, let's find a point where we can easily find the y value.

When x is equal to 4,5, or 6, there is an integer value for y. so let's plugin 5:

(5-3)(5-7)=-4

We can see when x is five, y has to be -12, so that is not the answer, thus d is the final answer.

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

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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).

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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

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

To convert billion to million you multiple by 1000

$P(0)' = 0.00833901 \:Billion/year \cdot 1000 = 8.34 \:Million/year \\P(1)' = 0.00838904 \:Billion/year \cdot 1000 = 8.39 \:Million/year$

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