Ask question

Ask question

All categories

3 years ago

8

Three angles of a regular pentagon are 60(degrees),120(degrees)and80(degrees).The remaining angles are congruent.Find the size o

f each of the remaining angles

1 answer:

mash [69]3 years ago

3 0

U would just add then together

You might be interested in

HELP PLS EASY POINTS (probably)

DIA [1.3K]

Answer:

C.

Step-by-step explanation:

3 0

3 years ago

What is the value of y=4x−2<br> when x=3<br> ?

MariettaO [177]

Answer:

y = 10

Step-by-step explanation:

y = 4(3) - 2

y = 12 - 2

y = 10

7 0

4 years ago

Read 2 more answers

Can Someone please help me?

SOVA2 [1]

Answer:

?????????????????????

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:

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

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

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

4 0

3 years ago

Read 2 more answers

Find the indicated limit, if it exists. limit of f of x as x approaches 0 where f of x equals 5x minus 8 when x is less than 0 a

vlabodo [156]

The limit does not exist. Why? Because the left hand limit DOES NOT equal the right hand limit. Let’s double check:

We could use -0.000001 to represent the left hand limit. This is less than 0. We plug in 5x - 8

5(-0.000001) - 8
-0.000005 - 8
-8.000005
If we would continue the limit (extend the zeros to infinity), we would get exactly
-8
That is our left hand limit.

Our right hand limit will be represented by 0.000001. This is greater than 0. We plug in abs(-4 - x)

abs(-4 - (0.000001))
abs(-4.000001)
4.000001
If we would continue the limit (extend the zeros to infinity), we would get exactly
4

4 does not equal -8, therefore

The limit does not exist

4 0

3 years ago