That's impossible. Are you factoring?
![▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪](https://tex.z-dn.net/?f=%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%20%20%7B%5Chuge%5Cmathfrak%7BAnswer%7D%7D%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA)
The required equation is ~
![\boxed{ \sf{y = \frac{x}{2} + 2 }}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Csf%7By%20%3D%20%20%5Cfrac%7Bx%7D%7B2%7D%20%2B%202%20%7D%7D)
![\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}](https://tex.z-dn.net/?f=%20%5Clarge%20%5Cboxed%7B%20%5Cmathfrak%7BStep%5C%3A%5C%3A%20By%5C%3A%5C%3AStep%5C%3A%5C%3AExplanation%7D%7D)
Let's find the slope of given line ~
comparing it with general slope - intercept form of line (y = mx + c) we get, m = -2 (that is slope of the line)
let the slope of the required line be n
And, now since the required line Is perpendicular to the given line. the product of their slopes is -1
that is ~
slope of required line is ~ 1/2
now, let's use the point - slope form of line to find the equation of required (perpendicular) line (using point (0 , 2) ~
that is ~
here, m = slope ~
I hope it helps ~
Answer:
The answer to your question is below
Step-by-step explanation:
Use trigonometric functions to find x
13. cos 40 = 5/x
x = 5 / cos 40
x = 5 / 0.766
x = 6.5
14. tan 25 = x/9
x = 9tan 25
x = 9(0.466)
x = 4.2
15.- cos 65 = x/12
x = 12 cos 65
x = 12 (0.423)
x = 5.1
16.- sin 32 = 15/x
x = 15 / sin 32
x = 15 / 0.53
x = 28.3
17.- tan 55 = 20 / x
x = 20 / tan 55
x = 20 / 1.42
x = 14
18.- sin31 = x / 18
x = 18 sin 31
x = 18 (0.52)
x = 9.3
Answer: 5
Step-by-step explanation: itswhere the 2 lines meet at the X