Answer: The distance of Observer A from the radio antenna is what fraction of the distance of Observer B from the radio antenna?
It is 1/4.
Step-by-step explanation:
We know that the intensity of electromagnetic waves decreases with the radius squared, this means that we can write a simple relation as:
Intensity(r) = A/r^2
Observer A measures 16 the intensity of observer B.
if Ia is the intensity that observer A measures and Ib is the intensity that observer B measures, we have that:
Ia = 16Ib
A/(ra)^2 = 16*A/(rb)^2
1/(ra)^2 = 16/(rb)^2
rb^2 = 16*ra^2
and we know that 16 = 4*4 = 4^2
rb^2 = (4*ra)^2
then rb = 4*ra
this means that the distance between observer B and the antenna is equal to 4 times the distance between observer A and the antenna.
The fraction is ra = rb/4
The distance of
Observer A from the radio antenna is what fraction of the distance of Observer B from the radio antenna?
It is 1/4.
Answer:
The answer can be explained here: theraleighregister.com/the-diagram-shows-the-width-and-area-of-a-rectangl.html
Step-by-step explanation:
It is best to make a table for this one
Hour | Cells
0 | 1
1 | 2
2 | 4
3 | 8
4 | 16
5 | 32
6 | 64
7 | 128
After 7 hours, there are 128 cells. To find this, you multiply the starting amount by two each hour.
Answer:
<h2>x = 9, y = -72/5</h2>
Step-by-step explanation:
Given the 2 by 2 matrix ![A = \left[\begin{array}{-cc}x&5\\y&-8\\\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7B-cc%7Dx%265%5C%5Cy%26-8%5C%5C%5Cend%7Barray%7D%5Cright%5D)
, we are to find the value of x and y for the expression A² = A to be true.
First we need to find A² by multiplying the same matrix together
![A^2 = \left[\begin{array}{-cc}x&5\\y&-8\\\end{array}\right] \left[\begin{array}{-cc}x&5\\y&-8\\\end{array}\right]\\\\ \ we \ normally \ multiply \ the \ rows \ of \ the \ first \ matrix \ with \ the \ column \ of \ the \ second \\\\A^2 = \left[\begin{array}{-cc}x^2+5y&5x-40\\xy-8y&5y+64\\\end{array}\right]\\\\Since \ A^2 = A,\ hence;\\\\\left[\begin{array}{-cc}x^2+5y&5x-40\\xy-8y&5y+64\\\end{array}\right] = \left[\begin{array}{-cc}x&5\\y&-8\\\end{array}\right]\\](https://tex.z-dn.net/?f=A%5E2%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7B-cc%7Dx%265%5C%5Cy%26-8%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7B-cc%7Dx%265%5C%5Cy%26-8%5C%5C%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%20%5C%20we%20%5C%20normally%20%5C%20multiply%20%5C%20the%20%5C%20rows%20%5C%20of%20%5C%20the%20%5C%20first%20%5C%20matrix%20%5C%20with%20%5C%20the%20%5C%20column%20%5C%20of%20%5C%20the%20%5C%20second%20%5C%5C%5C%5CA%5E2%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7B-cc%7Dx%5E2%2B5y%265x-40%5C%5Cxy-8y%265y%2B64%5C%5C%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5CSince%20%5C%20A%5E2%20%3D%20A%2C%5C%20hence%3B%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7B-cc%7Dx%5E2%2B5y%265x-40%5C%5Cxy-8y%265y%2B64%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7B-cc%7Dx%265%5C%5Cy%26-8%5C%5C%5Cend%7Barray%7D%5Cright%5D%5C%5C)
Equating the first row and second column of both matrices together, we will have;
5x-40 = 5
add 40 to both sides of the equation
5x-40+40 = 5+40
5x = 45
x = 45/5
x = 9
Similarly, we will equate the second row and second column of both matrices to have;
5y+64 = -8
Subtract 64 from both sdies
5y+64-64 = -8-64
5y = -72
y = -72/5
<em>Hence the value of x is 9 and y is -72/5</em>
