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uranmaximum [27]
4 years ago
8

Solve using the substitution and elimination method 3a-12b=9 4a-5b=3 can someone please help?

Mathematics
2 answers:
Veronika [31]4 years ago
7 0
3a-12b=9 => a = 4b + 3; but, <span>4a-5b=3 => 4(4b + 3) -5b = 3 => 11b = -9 => b = -9/11 => a = -36/11 + 3 = -3/11;</span>
tensa zangetsu [6.8K]4 years ago
3 0

Answer:6969

Step-by-step explanation:

Its the best answer

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What is the range of the function y=2x+1 for the domain 2&lt;=x&lt;=5?
egoroff_w [7]
So domain is the number you can use
range is the output your get from inputting the domain given
so from 2≤x≤5
since it is linear, we can be sure that we only need to test the endpoints of the domain to find the endpoints of the range

sub 2 for x
y=2(2)+1
y=4+1
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sub 5 for x
y=2(5)+1
y=10+1
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in interval notation: [5,11]
in other notation 5≤y≤11
or
R={y|5≤y≤11}
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4 years ago
Write 9/4 as a decimal. Round to three decimal places
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Read 2 more answers
adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the
Zigmanuir [339]

Answer:

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659 and \left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

Step-by-step explanation:

The equation of the isotope decay is:

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

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

\tau = \frac{5568\,years}{\ln 2}

\tau \approx 8032.926\,years

The decay time is:

t = 1315\,years + 2007\,years \pm 13\,years (There is no a year 0 in chronology).

t = 3335 \pm 13\,years

Lastly, the relative amount is estimated by direct substitution:

\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659

\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

4 0
3 years ago
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