Step-by-step explanation:
option C is correct
as,
4x is a common magnification for scanning objectives and, when combined with the magnification power of a 10x eyepiece lens, a 4x scanning objective lens gives a total magnification of 40x.
<em><u>hope </u></em><em><u>this </u></em><em><u>answer </u></em><em><u>helps </u></em><em><u>you </u></em><em><u>dear!</u></em><em><u> </u></em><em><u>take </u></em><em><u>care </u></em><em><u>and </u></em><em><u>may </u></em><em><u>u </u></em><em><u>have</u></em><em><u> a</u></em><em><u> great</u></em><em><u> day</u></em><em><u> ahead</u></em><em><u>!</u></em>
The equation of the vertical line passing through point -4,7 would be X = -4.
Answer: C) tan(pi/56)
=============================================
Explanation:
I recommend using a trig identity reference sheet. The specific identity we will be using is ![\frac{\tan(A)-\tan(B)}{1+\tan(A)\tan(B)} = \tan(A-B)](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctan%28A%29-%5Ctan%28B%29%7D%7B1%2B%5Ctan%28A%29%5Ctan%28B%29%7D%20%3D%20%5Ctan%28A-B%29)
What we are given is in the form
with A = pi/7 and B = pi/8
A-B = (pi/7)-(pi/8)
A-B = pi(1/7-1/8)
A-B = pi(8/56 - 7/56)
A-B = pi*(1/56)
A-B = pi/56
Therefore,
![\frac{\tan\left(\pi/7\right)-\tan(\pi/8)}{1+\tan(\pi/7)\tan(\pi/8)} = \tan\left(\pi/56\right)](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctan%5Cleft%28%5Cpi%2F7%5Cright%29-%5Ctan%28%5Cpi%2F8%29%7D%7B1%2B%5Ctan%28%5Cpi%2F7%29%5Ctan%28%5Cpi%2F8%29%7D%20%3D%20%5Ctan%5Cleft%28%5Cpi%2F56%5Cright%29)
Answer:
10 times.
Step-by-step explanation:
The 1 in the 615 is in the tenths place.
The 1 in the 91 is in the ones place.
So, the one in the 615 is 10 times greater than the one in 91.
Look:
1x10 is 10, the 1 in 615 is basically 10, and the 1 in 91 is basically 1.