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3 years ago

Cobalt-60 is used in the treatment of cancer. A sample contains 8 grams of co-60 wbere k=0.1308

1 answer:

5 0

No idea sorry <span>Cobalt-60 is used in the treatment of cancer. A sample contains 8 grams of co-60 wbere k=0.1308</span>

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Browns department store is having a 1/4 off sale. How much would a shirt that regularly sells for 36.00 cost during the sale?

nadya68 [22]

Since the shirt is 25 percent off your paying for 75 percent of the shirt. 0.75(36) is 27. The shirt will cost 27.00

4 0

3 years ago

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Mashcka [7]

Multiplying eqn no. two with 4 we add it up with eqn 1.

$4x - 5y = 10 \\ - 4x - 12y = 32 \\$

$- 17y = 42 \\ y = \frac{ - 42}{17}$

When

$- x - 8 = 3y \\ x + 3y = - 8$

$x = - 8 - 3 \frac{ - 42}{17 } \\ x = - 8 + \frac{42}{17} \\ x = \frac{ - 10}{17}$

So, x = - 10/17 and y = - 42/17.

5 0

3 years ago

Read 2 more answers

1. In a 30-60-90 triangle, the length of the hypotenuse is 6. What is the length of the shortest side?

il63 [147K]

Answer:

i think it’s b

6 0

2 years ago

Help ASAP please.......

Sergeeva-Olga [200]

Answer:

Its hard to tell from there being no graphing lines, however i believe the answer is 4, please let me know if the answer was correct

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)

Triss [41]

$g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}$

The surface element is

$\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

and the integral is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)$

###

To compute the last integral, you can integrate by parts:

$u=r\implies\mathrm du=\mathrm dr$

$\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}$

$\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr$

For this integral, consider a substitution of

$r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds$

$\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds$

$\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds$

$=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds$

and the result above follows.

4 0

3 years ago