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2 years ago

What 6x3 and 3x6 thankkkkkkkkkkkkkkkkk

Mathematics

2 answers:

Fed [463]2 years ago

8 0

Answer:

If you meant by (6x)³ and (3x)⁶.

(6x)³ = 6x · 6x · 6x = 108x³

(3x)⁶ = 3x · 3x · 3x · 3x · 3x · 3x = 729x⁶

If you meant by 6x³ and 3x⁶, then it is as it is.

If you meant 6 × 3 and 3 × 6,

6 × 3 = 18

3 × 6 = 18

6 × 3 = 3 × 6

yanalaym [24]2 years ago

5 0

Answer 6x3=18 and 3x6=18

Step-by-step explanation:

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What is the Lcm of 2 and 7

wel

Answer:

Step-by-step explanation:

The least common multiple is a number that both numbers can multiply something to get to, and the smallest possible common one.

Let's try listing out the multiples of 2:

2, 4, 6, 8, 10, 12, 14, 16, 18, 20

And the multiples of 7:

7, 14, 21, 28, 35, 42, 49, 56, 63, 70

Notice that both have the number 14 in common, and it is the first number that they do so. There are lots of other multiples that they have in common too, but 14 is the least.

5 0

2 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: