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torisob [31]
3 years ago
10

Radiation​ machines, used to treat​ tumors, produce an intensity of radiation that varies inversely as the square of the distanc

e from the machine. At 3​ meters, the radiation intensity is 62.5 milliroentgens per hour. What is the intensity at a distance of 2.7 ​meters?The intensity is______milliroentgens per hour. (Round to the nearest tenth as needed.)
Mathematics
1 answer:
iren [92.7K]3 years ago
4 0

Answer:

77.2 milliroentgens per hour

Step-by-step explanation:

The first step is to find the constant of proportionality

I ∝ 1/d²

I = k/d²

62.5 = k/(3)²

k = 562.5

The second step is to calculate the intensity at 2.7 meters:

I = 562.5/d²

I = 562.5/(2.7)²

I = 77.2 milliroentgens per hour

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erma4kov [3.2K]

Answer:

hope its right lol

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3 years ago
Use &lt;, &gt;, or = to fill in the blanks.<br> -7----- 0<br> Please help me!
Oduvanchick [21]

Answer:

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Step-by-step explanation:

3 0
3 years ago
What is the answer to:
velikii [3]
Weird way to write it but alright! (Sideways) 

19pq^-2 x 5pq^6 = ?

These problems are pretty much single operations between each of the variables / constants. 

So it's like this: 

(19*5)(p*p)(q^-2*q^6) = ?

19*5 is 95.

For p*p remember that when two variables multiply there given powers add. In the case where the powers are not shown (like in the case of p*p) they are always assumed to be 1. So what is 1+1? 2. 

p*p is p^2

For q^-2*q^6 it is the same deal with the previous problem. So now the problem looks like this: 

-2 + 6 = 4 
(The two is negative, because the power is negative 2) 

So, q^4. 

Our final answer is all of the combined.... like a so: 

95p^2q^4
3 0
3 years ago
Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab
Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx

If f(x)=x^2, then

f'(x)=2x

and

b=0\\ \\a=2

Therefore,

SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx

Apply substitution

x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du

Then

SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du

Now

\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))

Hence,

SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}

3 0
3 years ago
Can someone help me and explain on paper , I’m still so confused
Oksanka [162]

Answer:

Your answer is -5

Step-by-step explanation:

On paper;

8 0
3 years ago
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