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2 years ago

12

Is a/b +b/c rational number

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

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

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

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

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8 0

3 years ago

Help i need help please

lubasha [3.4K]

Answer: C

Step-by-step explanation:

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1/3 over the y axis.

7 0

2 years ago

Look at the factors of 24 and 32.

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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