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4 years ago

HELP!

1 answer:

Studentka2010 [4]4 years ago

4 0

So one way we can do this is-

2(2n+7)+3n=

4n+14+3n=

7n+14

Another simpler way possible is-

2n+7+2n+7+3n=

RE-ORDER

2n+2n+3n+7+7=

COMBINE LIKE TERMS

7n+14

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Solve the following system by graphing.

Masja [62]

Hello :
the solution is : <span>(3, -1)
put x= 3 and y = -1
you have : 3-(-1) = 4 and 3+(-1) = 2 ..... right</span>

7 0

3 years ago

Read 2 more answers

Friday night, a pizza parlor sold

SpyIntel [72]

The total is. Friday 573

6 0

4 years ago

The 2 in the middle is the one that I needed help <br> Please

r-ruslan [8.4K]

Answer:

The equation is p=34a+12, and a=7

Step-by-step explanation:

Perimeter is all of the sides added together

p=12a+6a+8+6a+8+10a-4

Combine like-terms

p=34a+12

Solve for a by making p equal to 250

250=34a+12

Subtract 12 from both sides

238=34a

Divide both sides by 34

7=a

4 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: