Use the commutative and associative properties as needed to simplify the expression. (12+a)+14

Can anyone help with only #1 please, will give you brainliest!!

Tresset [83]

No, it is <u>not</u> a translation. The arrow starts pointing to the right and then points upward. A translation would have the arrow always pointing right the entire time, only that its position would move.

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Problem 2

We can shift the K up 6 units to get what you see in the attached image below

5 0

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

Please help this is algebra 1<br> problem is in the picture

Naddika [18.5K]

Answer:

0,1,2,3,4,5

Step-by-step explanation:

You can not buy more than 5 books because you'd have a negative amount of money. So 0,1,2,3,4,5 are the possible values for b.

7 0

3 years ago

John can mow a lawn in 80 minutes. Rocky can mow the same lawn in 120 minutes. How long does it take for both John and Rocky to

mario62 [17]

Answer: it will take them 48 minutes.

Step-by-step explanation:

John can mow a lawn in 80 minutes. This means that the rate at which he moans the lawn per minute is 1/80

Rocky can mow the same lawn in 120 minutes. This means that the rate at which Rocky can mow the same lawn per minute is 1/120

If they work together, they would work simultaneously and their individual rates are additive. This means that their combined working rate would be

1/80 + 1/120 = 200/9600 = 1/48

Assuming it takes t hours for both of them to clean the room working together, the working rate per hour would be 1/t. Therefore,

1/48 = 1/t

t = 48 minutes

4 0

3 years ago

An element with mass 670 grams decays by 27.3% per minute. How much of the element is remaining after 9 minutes, to the nearest

ArbitrLikvidat [17]

There will be 38 grams remaining.

The equation would be of the form
y = a(1+r)ˣ, where a is the initial value, r is the rate as a decimal number, and x is the amount of time. Using our values from the problem, we have:

y = 670(1-0.273)^9 = 670(0.727)^9 = 38

6 0

3 years ago

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