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marissa [1.9K]
3 years ago
13

Need help ASAP please please

Mathematics
2 answers:
natali 33 [55]3 years ago
5 0
Open end arrow pointing to the right from -9 to past -2

For all x >-9 is any number on the number line above, to the right of -9
Rama09 [41]3 years ago
4 0

Answer:

-10,-11, -12

Step-by-step explanation:

All are bigger

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Find the area of a circle with a circumference of 12.56 units​
Fudgin [204]
Answer: The area is 12.55

8 0
3 years ago
What is the interest earned on $20,000 for five years, at an interest rate of 3% compounded daily?
SSSSS [86.1K]

Answer:

The final balance is $23,232.33.

The total compound interest is $3,232.33.

6 0
2 years ago
12=5x-13x-44. answer it in multi-step equations
fomenos

12=5x-13x-44

combine like terms

12 = -8x -44

add 44 to each side

12+44 = -8x

56= -8x

divide by -8 on each side

56/-8 = x

-7 =x


5 0
3 years ago
Read 2 more answers
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
1. 4x– 15 = 17 - 4x+10x<br> Help?
Nadya [2.5K]

Answer:

x=-16

Step-by-step explanation:

4x– 15 = 17 - 4x+10x

4x+4x-10x = 17+15

-2x=32

-x=16

x=-16

7 0
3 years ago
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