Answer:
diamerer of the hair (t) 3.11 X 10^{-5} m
Step-by-step explanation:
The first dark spot is at the zero location where the glass plates touch. To find the number of dark fringes after first dark fringe, divide 12 cm by 0.75 mm (.075 cm)
12/.075 = 160.0
total number of additional bright fringe is 156 because there is no further portion for extra dark fringe.
diamerer of the hair (t) can be derived as
2nt = (m + .5)*wavelength
2(1.52)(t) = (160.5)(589.3 X 10^{-9})
t = 3.11 X 10^{-5} m
Yes. Conceptually, all the matrices in the group have the same structure, except for the variable component 
. So, each matrix is identified by its top-right coefficient, since the other three entries remain constant.
However, let's prove in a more formal way that
![\phi:\ \mathbb{R} \to G,\quad \phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cphi%3A%5C%20%5Cmathbb%7BR%7D%20%5Cto%20G%2C%5Cquad%20%5Cphi%28x%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20)
is an isomorphism.
First of all, it is injective: suppose 
. Then, you trivially have 
, because they are two different matrices:
![\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right],\quad \phi(y) = \left[\begin{array}{cc}1&y\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cphi%28x%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%2C%5Cquad%20%5Cphi%28y%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26y%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20)
Secondly, it is trivially surjective: the matrix
![\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cphi%28x%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20)
is clearly the image of the real number x.
Finally, 
and its inverse are both homomorphisms: if we consider the usual product between matrices to be the operation for the group G and the real numbers to be an additive group, we have
![\phi (x+y) = \left[\begin{array}{cc}1&x+y\\0&1\end{array}\right] = \left[\begin{array}{cc}1&x\\0&1\end{array}\right] \cdot \left[\begin{array}{cc}1&y\\0&1\end{array}\right] = \phi(x) \cdot \phi(y)](https://tex.z-dn.net/?f=%20%5Cphi%20%28x%2By%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%2By%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26y%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cphi%28x%29%20%5Ccdot%20%5Cphi%28y%29)
<span>There are 4 vans. So we have that probability that the first vehicle is a van p (e) = 4/10 = 0.4.
P(e|f) = P( f and e) / p (f)
p(e) = 0.4 and p(f) = 3/9
P (f and e) = 0.40 * 0.33 = 0.132
So p(e|f) = 0.4 * 0.33/ 0.33 = 0.4
P(f and e) p(f) * p(e) = 0.4 * 0.33 = 0.132</span>
To find the y-intercept, let x = 0.
0 - 3y = -6
-3y = -6
y = 2
The y-intercept of the equation is the point (0, 2).
Since you don't want any leftover items, you have to find the greatest common factor (GCF) of both the crayons and bubbles. Since the highest common factor for both 28 and 12 is 4, the greatest number of gift bags you can make without any leftover items would be 4 gift bags.
Hope I could help.
