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3 years ago

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Mathematics

1 answer:

galina1969 [7]3 years ago

5 0

Answer:

Step-by-step explanation:

It would be 72. You get that by doing 12 X 12 which is 144. Than divide it by 2 to get 72.

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Find the area of the right triangle with a base of 38cm and the height of 26cm

sesenic [268]

The answer is D!

Area for a triangle solution is 1/2(b•h)
Insert each number!
1/2(38•26)
Area = 494cm

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3 years ago

What are 3 equivalent fractions of 5/8

antoniya [11.8K]

Answer:

Here are the three equivalent fractions of 5/8.

10/16 15/24 20/32

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Which of the following shows the least expensive unit price?

Arlecino [84]

It would be the first one because each orange costs $0.34

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David invests 10,000 in a sqving account that pays 3.5% simple interest. If David makes no withdrawals or deposits to the accoun

Triss [41]

Answer:

Step-by-step explanation:

The formula for determining simple interest is expressed as

I = PRT/100

Where

I represents interest paid on the amount invested.

P represents the principal or amount invested.

R represents interest rate

T represents the duration of the investment in years.

From the information given,

P = 10000

R = 3.5%

T = 7 years

I = (10000 × 3.5 × 7)/100 = $2450

The amount that will be in the account after 7 years is

10000 + 2450 = $12450

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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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3 years ago

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