![tan \frac{x}{2} =\pm \sqrt{\frac{1-cos x}{1+cos x}}](https://tex.z-dn.net/?f=tan%20%5Cfrac%7Bx%7D%7B2%7D%20%3D%5Cpm%20%5Csqrt%7B%5Cfrac%7B1-cos%20x%7D%7B1%2Bcos%20x%7D%7D)
Find cos using trig identities:
![sec x = \frac{1}{cos x} \\ tan^2 x = sec^2 x -1](https://tex.z-dn.net/?f=sec%20x%20%3D%20%5Cfrac%7B1%7D%7Bcos%20x%7D%20%20%5C%5C%20tan%5E2%20x%20%3D%20sec%5E2%20x%20-1)
Therefore
![cos x = \frac{1}{sec x} =\pm \frac{1}{\sqrt{tan^2 x +1}}](https://tex.z-dn.net/?f=cos%20x%20%3D%20%5Cfrac%7B1%7D%7Bsec%20x%7D%20%3D%5Cpm%20%5Cfrac%7B1%7D%7B%5Csqrt%7Btan%5E2%20x%20%2B1%7D%7D)
Sub in tan x = 3, (Note that x is in 3rd quadrant, cos x < 0)
![cos x =- \frac{1}{\sqrt{3^2 +1}} = -\frac{1}{\sqrt{10}}](https://tex.z-dn.net/?f=cos%20x%20%3D-%20%5Cfrac%7B1%7D%7B%5Csqrt%7B3%5E2%20%2B1%7D%7D%20%3D%20-%5Cfrac%7B1%7D%7B%5Csqrt%7B10%7D%7D)
Finally, sub into Half-angle formula:(Note x/2 is in 2nd quadrant, tan x<0)
Do it step by step
ab+3c
(-6)(-2)+3(5)
A negative times a negative is positive. (-)(-)=+
12+3(5)
12+15
=27
Answer:
m=2 or y=2x+1
Step-by-step explanation:
first use the y2-y1/x2-x1 formula and then use y=mx+b formula
plug in 3 for y2 and 1 for y1
plug in 1 for x2 and 0 for x1
![\frac{3-1}{1-0} =\frac{2}{1} =2](https://tex.z-dn.net/?f=%5Cfrac%7B3-1%7D%7B1-0%7D%20%3D%5Cfrac%7B2%7D%7B1%7D%20%3D2)
so ur slope is 2
_______________
for y=mx+b *extra info ig if u need it just incase*
i usually use the first point for this but u can use any one u want
plug in 0 for x
plug in 1 for y
and plug in 2 for m
1=2(0)+b
2*0=0
1=0+b
subtract o on both sides
1=b and b is ur y-intercept
y=2x+1
hope this helps
You must convert the percentage to decimal notation and then multiply it by 72.
As the name implies, percentage is a something per one-hundred; so to convert any percentage to decimal notation we divide by one-hundred:
![\frac{310}{100}=3.1](https://tex.z-dn.net/?f=%20%5Cfrac%7B310%7D%7B100%7D%3D3.1%20)
So the 310% of 72 is:
Answer:
The slope or gradient of a line is a number that describes both the direction and the steepness of the line.