Answer:
i made this poem
Step-by-step explanation:
Send this to anyone that feels left out or you love please.
No Need to be a shadow i see you
No need to be muted i can hear you
No need to cut your not paper
No need to Hang your not a coat
No need to Fall apart because of a guy you'll soon find better
No need to play with your life You are not a toy
No need to not feel loved I love you
No need to feel Numb I'm here
No need to pop pills Your loved
No need to Commit suic.ide There is to much to life
I think all I'm saying is Life is Way more then school, job, rest and repeat There's adventures Happy moments and Horrifying times But we all soon get over these things Your in this world in a life For a reason Life it with a purpose
Answer: y = 4x/3 - 5/2
Step-by-step explanation:
The equation of a straight line can be represented in the slope intercept form as
y = mx + c
Where
c represents the y intercept
m = slope = (y2 - y1)/(x2 - x1)
The given line, L1 passes through A(6, - 7) and B(- 6, 2). The slope of line L1 is
m = (2 - - 7)/(- 6 - 6) = 9/ -12 = - 3/4
If two lines are perpendicular, it means that the slope of one line is the negative reciprocal of the slope of the other line.
Therefore, the slope of line L2 passing through the midpoint, M is 4/3
The formula determining the midpoint of a line is expressed as
[(x1 + x2)/2 , (y1 + y2)/2]
Midpoint, M = [(6 + -6)/2 , (- 7 + 2)/2]
= (0, - 5/2]
This means that the y intercept of line L2 is - 5/2
The equation of L2 becomes
y = 4x/3 - 5/2
They are vertical angles, thus they are equal to each other:
5x + 12 = 6x - 10
22 = x
x = 22
Answer:
Equation of tangent plane to given parametric equation is:
Step-by-step explanation:
Given equation
---(1)
Normal vector tangent to plane is:
Normal vector tangent to plane is given by:
![r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]](https://tex.z-dn.net/?f=r_%7Bu%7D%20%5Ctimes%20r_%7Bv%7D%20%3Ddet%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Chat%7Bi%7D%26%5Chat%7Bj%7D%26%5Chat%7Bk%7D%5C%5Ccos%28v%29%26sin%28v%29%260%5C%5C-usin%28v%29%26ucos%28v%29%261%5Cend%7Barray%7D%5Cright%5D)
Expanding with first row
at u=5, v =π/3
---(2)
at u=5, v =π/3 (1) becomes,
From above eq coordinates of r₀ can be found as:
From (2) coordinates of normal vector can be found as
Equation of tangent line can be found as:
