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FrozenT [24]
2 years ago
12

Is a/b +b/c rational number​

Mathematics
1 answer:
Anarel [89]2 years ago
3 0

Answer:

Yes.

Step-by-step explanation:

Yes.

Assuming a, b and c are integers (not = 0)

a/b + b/c

= (ac + b^2) / bc   which is a rational number.

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Find the length of the curve. R(t) = 7t, 3 cos(t), 3 sin(t) , −2 ≤ t ≤ 2
Leviafan [203]

we are given

r(t)=(7t,3cos(t),3sin(t))

we can find x , y and z

x=7t,y=3cos(t),z=3sin(t))

now, we can use arc length formula

L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt

now, we can find derivative

x'=7,y'=-3sin(t),z'=3cos(t))

now, we can plug values

and we get

L=\int _{-2}^2\sqrt{7^2+\left(-3\sin \left(t\right)\right)^2+\left(3\cos \left(t\right)\right)^2}dt

L=\int _{-2}^2\sqrt{7^2+9}dt

=\sqrt{58}\cdot \:2-\left(-2\sqrt{58}\right)

L=4\sqrt{58}...........Answer


8 0
3 years ago
Help i need help please
lubasha [3.4K]

Answer: C

Step-by-step explanation:

On the big triangle, the length from the origin to point A is three blocks. On the new triangle it is only one block. And it is getting smaller, so it is a fraction.

1/3 over the y axis.

7 0
2 years ago
Look at the factors of 24 and 32.
Salsk061 [2.6K]

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4 0
2 years ago
Read 2 more answers
The ​half-life of a radioactive element is 130​ days, but your sample will not be useful to you after​ 80% of the radioactive nu
gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

\frac{dN}{dt}\propto -N

\Rightarrow \frac{dN}{dt} =-\lambda N

\Rightarrow \frac{dN}{N} =-\lambda dt

Integrating both sides

\int \frac{dN}{N} =\int\lambda dt

\Rightarrow ln|N|=-\lambda t+c

When t=0, N=N_0 = initial amount

\Rightarrow ln|N_0|=-\lambda .0+c

\Rightarrow c= ln|N_0|

\therefore ln|N|=-\lambda t+ln|N_0|

\Rightarrow ln|N|-ln|N_0|=-\lambda t

\Rightarrow ln|\frac{N}{N_0}|=-\lambda t.......(1)

                            \frac{N}{N_0}=e^{-\lambda t}.........(2)

Logarithm:

  • ln|\frac mn|= ln|m|-ln|n|
  • ln|ab|=ln|a|+ln|b|
  • ln|e^a|=a
  • ln|a|=b \Rightarrow a=e^b
  • ln|1|=0

130 days is the half-life of the given radioactive element.

For half life,

N=\frac12 N_0,  t=t_\frac12=130 days.

we plug all values in equation (1)

ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130

\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130

\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130

\rightarrow -ln|2|=-\lambda \times 130

\rightarrow \lambda= \frac{-ln|2|}{-130}

\rightarrow \lambda= \frac{ln|2|}{130}

We need to find the time when the sample remains 80% of its original.

N=\frac{80}{100}N_0

\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t

\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t

\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t

\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}

\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}

\Rightarrow t\approx 42

We can use the sample about 42 days.

7 0
3 years ago
What’s 6 - 7/15 in the simplest form
NeX [460]

90/15-7/15=83/15 the correct answer to this equation.

4 0
3 years ago
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