So I have to angles a
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Answer:
Step-by-step explanation:
for two lines to be parallel they MUST have the same slope but have different Y intercepts ( Y axis crossing values)
parallel lines will never intersect each other
first I use a graphing calcuator to solve the problem
I like to use y intercept form better, but you don't have use it
I will try to solve it directly without going through the y intercept equation
x + 5y = 10 in y intercept form y = mx + b is
5y = -x + 10 divide both sides by 5
y = -x/5 + 10/5
y = -x/5 + 2
Now I need to find a parallel line that passes through the x =1 and y = 3 point recall this line will have the same slope = m = -1/5
and a different Y axis crossing point 'b' that we don't know
y = -x/5 + b y = 3 when x = 1 so slove for 'b'
3 = -1/5 + b
3 = -0.2 + b add -0.2 to both sides
3.2 = b
y = -x/5 + 3.2 this is answer in y intercept form
if you multiply both sides by 5
5y = -x + 16 add x to both sides
x + 5y = 16 answer in standard form
THE EASIER WAY TO SOLVE THE PROBLEM IS ......
x + 5y = 10 find a line parallel that passes through x = 1, y = 3
x + 5y = Z Z is an unknown value plug in x and y and solve for Z
1 + 5(3) = Z
Z = 16 so the parallel line in standard form is
x + 5y = 16
same as before, my y intercept method was correct, but not worth the effort
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——————————
You have y as an implicit function of x:
sin(xy) – x = 0
Use implicit differentiation. As y is a function of x, then you must apply the chain rule there:
d d
—— [ sin(xy) – x ] = —— (0)
dx dx
d d d
—— [ sin(xy) ] – —— (x) = —— (0)
dx dx dx
d
—— [ sin(xy) ] – 1 = 0
dx
d
—— [ sin(xy) ] = 1
dx
d
cos(xy) · —— (xy) = 1
dx
Now, apply the product rule for that last derivative:
![\mathsf{cos(xy)\cdot \left[\dfrac{d}{dx}(x)\cdot y+x\cdot \dfrac{dy}{dx}\right]=1}\\\\\\ \mathsf{cos(xy)\cdot \left[1\cdot y+x\cdot \dfrac{dy}{dx}\right]=1}\\\\\\ \mathsf{y\,cos(xy)+x\,cos(xy)\cdot \dfrac{dy}{dx}=1}](https://tex.z-dn.net/?f=%5Cmathsf%7Bcos%28xy%29%5Ccdot%20%5Cleft%5B%5Cdfrac%7Bd%7D%7Bdx%7D%28x%29%5Ccdot%20y%2Bx%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%5D%3D1%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7Bcos%28xy%29%5Ccdot%20%5Cleft%5B1%5Ccdot%20y%2Bx%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%5D%3D1%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7By%5C%2Ccos%28xy%29%2Bx%5C%2Ccos%28xy%29%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D%3D1%7D)
dy
Isolate —— :
dx
dy
x cos(xy) · —— = 1 – y cos(xy)
dx
Assuming x cos(xy) ≠ 0,
dy 1 – y cos(xy)
—— = ———————— <——— this is the answer.
dx x cos (xy)
I hope this helps. =)
——————————
You have y as an implicit function of x:
sin(xy) – x = 0
Use implicit differentiation. As y is a function of x, then you must apply the chain rule there:
d d
—— [ sin(xy) – x ] = —— (0)
dx dx
d d d
—— [ sin(xy) ] – —— (x) = —— (0)
dx dx dx
d
—— [ sin(xy) ] – 1 = 0
dx
d
—— [ sin(xy) ] = 1
dx
d
cos(xy) · —— (xy) = 1
dx
Now, apply the product rule for that last derivative:
dy
Isolate —— :
dx
dy
x cos(xy) · —— = 1 – y cos(xy)
dx
Assuming x cos(xy) ≠ 0,
dy 1 – y cos(xy)
—— = ———————— <——— this is the answer.
dx x cos (xy)
I hope this helps. =)