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3 years ago

Which of the following is most likely the next step in the series

Mathematics

2 answers:

Scorpion4ik [409]3 years ago

8 0

Answer:

The answer is B

Step-by-step explanation:

You have to find the pattern, in this case it's 1,3,5 and 7

Hope this helps :)

poizon [28]3 years ago

4 0

Answer:

Step-by-step explanation:

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Sever21 [200]

D
because first you times z and x to get the area and then you multiply that by two added to
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8 0

3 years ago

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Help please<br><br> 1 1/2% of what number is 10.05

MrRissso [65]

Answer:

10/25%

Step-by-step explanation:

8 0

3 years ago

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The graph models the daily high temperature, H(t), in Santiago, Chile, t days after the hottest day of the year. What does the p

alekssr [168]

For this case we have a function of the form:
H (t)
Where,
t: number of days
H: the daily high temperature
We have the point (0.29).
It means that the hottest day of the year had a temperature of 29 degrees.
Answer:
the hottest day of the year had a temperature of 29 degrees.

5 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: