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2 years ago

Please find the answer my math teacher doesn’t know how to explain!

Mathematics

1 answer:

const2013 [10]2 years ago

8 0

Answer:

68.8

Step-by-step explanation:

(a+b)h/2

12.9+8.6 *6.4/2

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2.4x-9 < 1.8x+6 solve the inequality

nasty-shy [4]

Answer:

x<25

Step-by-step explanation:

Let's solve your inequality step-by-step.

2.4x−9<1.8x+6

Step 1: Subtract 1.8x from both sides.

2.4x−9−1.8x<1.8x+6−1.8x

0.6x−9<6

Step 2: Add 9 to both sides.

0.6x−9+9<6+9

0.6x<15

Step 3: Divide both sides by 0.6

0.6x/0.6 < 15/0.6

x<25

8 0

2 years ago

Mane who be shaking dice?

skelet666 [1.2K]

Not me

Step-by-step explanation:

5 0

2 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: