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Alona [7]
3 years ago
9

A soduim isotpe, Na24, has a life of 15 hours

Mathematics
1 answer:
masha68 [24]3 years ago
4 0
What's the question?
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The measurement of the complement of an angle exceeds the measure of the angle by 25%. Find the measurement of the angle and its
tester [92]
Let the angles be A and C 

<span>Complimentary angles, therefor </span>
<span>A + C = 90 </span>

<span>The measurement of the complement of an angle exceeds the measure of the angle by 25% (you used a % sign so I'll use % instead of degrees) </span>
<span>C = 1.25A </span>

<span>Substitute </span>
<span>A + 1.25A = 90 </span>
<span>2.25A = 90 </span>
<span>A = 40º <----- </span>
<span>C = 50º <----- </span>
<span>------------------------------ </span>

<span>If you meant 25º </span>
<span>A + (A + 25) = 90 </span>
<span>2A + 25 = 90 </span>
<span>2A = 65 </span>
<span>A = 32.5º </span>
<span>C = 57.5º</span>
3 0
3 years ago
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
Need help please asap
svlad2 [7]

Answer:

A

Step-by-step explanation:

8 0
3 years ago
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Hi it is Appa what is your name
Lynna [10]

Answer:

names Alicia :)

8 0
3 years ago
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3) what is the sum <br> 7/4 (-1/5) <br><br> a)31/20 <br> b)39/20 <br> c)6/9 <br> d)-7/20
Lady bird [3.3K]
7/4 + (-1/5) = 7/4 - 1/5 = (35 - 4)/20 = 31/20
6 0
3 years ago
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