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S_A_V [24]
2 years ago
15

Help!! For first and second please

Mathematics
1 answer:
jenyasd209 [6]2 years ago
5 0

Answer:

it goes al the way up to 15 and it stops at 5 and the 8 makes the green house

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The polygons in each pair are similar find the scale factor smaller figure to the larger
GrogVix [38]

Answer:

Smaller Figure 3

Larger Figure 5

Scale factor = 15/25 = 3/5

3 0
2 years ago
Crystal is spinning the spinner at right and claims she has a good chance of having the spinner land on red at least once in thr
-BARSIC- [3]
Wouldn’t the answer be 1/3?
6 0
3 years ago
What is the answer to this inequality?<br> (x+7)&gt;80
Crank

Answer:

x > 73

Step-by-step explanation:

x+7 > 80

subtract 7 from both sides

x + 7 - 7 > 80 - 7

x > 73

-Chetan K

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:

\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|

\Delta t_{1/2} \approx 65.788\,days

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

4 0
3 years ago
Read 2 more answers
How to show my work for2+[4+(5x6)]
Setler [38]

Step-by-step explanation:

2+[4+(5*6)]

2+[4+30]

2+34

36

7 0
2 years ago
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