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2 years ago

Find the GCF for the list. - 6x^2, 21x^3

Mathematics

1 answer:

wolverine [178]2 years ago

6 0

Answer:

3x^2

Step-by-step explanation:

3x^2 can be factored out of each

-6x^2 = 3x^2(-2)

21x^3 = 3x^2(7x)

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How many 5/8 pound bags does it take to get 16 pounds of soil?

Sauron [17]

Answer:

Step-by-step explanation:

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2 years ago

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Only answer this if you know 1000% it is correct I need this to be correct

Serggg [28]

I believe that the answer would be B because if you put the numbers into the variable spots, this is what you get. For example, when you put -3 into A, because it is the one with the x, and 6 in for B, you would get the first part of the equation. Then, put 12 in for C, you would get the whole equation. Sorry if this is wrong or if I did a bad job explaining this. Good luck!

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3 years ago

1 square kilometre equals to 100000000000 square centimetres Express it in scientific notation

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3 years ago

Carbon-14 is a radioactive isotope that has a half-life of 5,730 years. Approximately how many years will it take for carbon-14

PilotLPTM [1.2K]

Answer:

<em>Carbon-14 will take 19,035 years to decay to 10 percent.</em>

Step-by-step explanation:

<u>Exponential Decay Function</u>

A radioactive half-life refers to the amount of time it takes for half of the original isotope to decay.

An exponential decay can be described by the following formula:

$N(t)=N_{0}e^{-\lambda t}$

Where:

No = The quantity of the substance that will decay.

N(t) = The quantity that still remains and has not yet decayed after a time t

$\lambda$ = The decay constant.

One important parameter related to radioactive decay is the half-life:

$\displaystyle t_{1/2}=\frac {\ln(2)}{\lambda }$

If we know the value of the half-life, we can calculate the decay constant:

$\displaystyle \lambda=\frac {\ln(2)}{ t_{1/2}}$

Carbon-14 has a half-life of 5,730 years, thus:

$\displaystyle \lambda=\frac {\ln(2)}{ 5,730}$

$\lambda=0.00012097$

The equation of the remaining quantity of Carbon-14 is:

$N(t)=N_{0}e^{-0.00012097\cdot t}$

We need to calculate the time required for the original amout to reach 10%, thus N(t)=0.10No

$0.10N_o=N_{0}e^{-0.00012097\cdot t}$

Simplifying:

$0.10=e^{-0.00012097\cdot t}$

Taking logarithms:

$ln 0.10=-0.00012097\cdot t$

Solving for t:

$\displaystyle t=\frac{log 0.10}{-0.00012097}$

$t\approx 19,035\ years$

Carbon-14 will take 19,035 years to decay to 10 percent.

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2 years ago

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For the sequence 3n - 1, generate the first three terms.

Hunter-Best [27]

3n - 1

When n = 1,

3n - 1 = 3*1 - 1 = 3 -1 = 2

When n = 2,

3n - 1 = 3*2 - 1 = 6 -1 = 5

When n = 3,

3n - 1 = 3*3 - 1 = 9 -1 = 8

Hence, the first three terms are 2, 5 and 8.

5 0

3 years ago