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nalin [4]
3 years ago
12

Which of the following is a solution of y> lXI - 6?

Mathematics
1 answer:
xeze [42]3 years ago
4 0

Answer:

C

Step-by-step explanation:

if you plug the coordinates into the function it is correct unlike the other options

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Fill the value of the expression 24 3/5 +4 ×(8 1/5-2)
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First off, we'll convert the mixed fractions to "improper",

\bf \stackrel{mixed}{24\frac{3}{5}}\implies \cfrac{24\cdot 5+3}{5}\implies \stackrel{improper}{\cfrac{123}{5}}
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\stackrel{mixed}{8\frac{1}{5}}\implies \cfrac{8\cdot 5+1}{5}\implies \stackrel{improper}{\cfrac{41}{5}}

\bf 24\frac{3}{5}\times \left( 8\frac{1}{5}-2 \right)\implies \cfrac{123}{5}\left( \cfrac{41}{5}-2 \right)\impliedby \mathbb{PEMDAS}
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\cfrac{123\cdot 31}{5\cdot 5}\implies \cfrac{3813}{25}\implies 152\frac{13}{25}
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3 years ago
Nickel uses 5 ounces of mushrooms 2 ounces of butter and 1/4 pint of cream to make mushroom soup find the ratio of the number of
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2 years ago
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

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

6 0
3 years ago
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