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2 years ago

Plz help lolololololol

Mathematics

1 answer:

kicyunya [14]2 years ago

8 0

Answer:

3.14

Step-by-step explanation:

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What is 28.5 rounded to the nearest one

kumpel [21]

Answer:

Step-by-step explanation:

anythin0.5 or higher you round up so for 28.5 you round up to 29.

hope this helps. :)

8 0

3 years ago

Read 2 more answers

Please help ASAP!!!!!!!!!!!

Sergeeva-Olga [200]

Answer:

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3 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

Use the formula V = s³, where V is the volume and s is the edge length of the cube, to solve this problem. A cube-shaped box has

aev [14]

Using the given equation and edge length:

V = 45^3

V = 45 x 45 x 45

V = 91,125 cubic meters

3 0

4 years ago

Solve for g.<br> g/6 - -53 = 57

Mashcka [7]

Step-by-step explanation:

g/6 - -53 = 57

g/6 + 53 = 57

g/6 = 4

g = 4×6 = 24

5 0

3 years ago