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2 years ago

What is the sum 2, 3, 4, and 5?

Mathematics

1 answer:

borishaifa [10]2 years ago

4 0

Answer:

Step-by-step explanation:

We can find the sum of these numbers by adding them together.

The cumulative rule of addition lets us add numbers together in any order we want.

So, for 2 + 3 + 4 + 5 …

2 + 3 = 5

So now we have 5 + 4 + 5

5 + 4 = 9

Now we have 9 + 5

And 9 + 5 = 14

So, your answer is 14

Hope this helps!

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algol13

118,813,760 different combination license plates can the country produce

<h3 /><h3>permutations and combinations:</h3>

permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor. By considering the ratio of the number of desired subsets to the number of all possible subsets for many games of chance in the 17th century, the French mathematicians Blaise Pascal and Pierre de Fermat gave impetus to the development of combinatorics and probability theory.

given that

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3 0

1 year ago

Solve -9 2/7 - (-10 3/7).<br> A. - 1 1/7<br><br> B. 1 1/7<br><br> C. 19 1/7<br><br> D. 19 5/7

jenyasd209 [6]

Answer:

1 5/7

Step-by-step explanation:

-9 2/7 -(-10 3/7) recall -×- = +

-9 2/7 +10 3/7

1 2+3/7

1 5/7

6 0

3 years ago

Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot

Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

$\frac{dm}{dt} = -\frac{t}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in kilograms.

$t$ - Time, measured in days.

$\tau$ - Time constant, measured in days.

The solution of the differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

$\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}$

$\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }$

The half-life of a isotope ( $t_{1/2}$ ) as a function of time constant is:

$t_{1/2} = \tau \cdot \ln2$

$t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2$

The half-life difference between isotope B and isotope A is:

$\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|$

If $\frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9$ , $t_{A} = 33\,days$ and $t_{B} = 43\,days$ , the difference in the half-lives of the isotopes is: