Parametric equations
transformations : let x = (2 cos t) + h and y = (2 sin t) + k
(a) writing to learn : let k=0 and h= -2,-1,1, and 2, in turn. graph
using the parameter interval [0,2(pi)]. describe the role
ofh.
(b) writing to learn : let h=0 and k= -2,-1,1, and 2, in turn. graph
using the parameter interval [0,2(pi)]. describe the role
ofk.
(c) find a parametrization for the circle with radius 5 and
centerat (2,-3)
(d) find a parametrization for the ellipse centered at (-3,4)
withsemimajor axis of length 5 parallel to the x-axis and
semiminoraxis of length 2 parallel to the y-axis.

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.

step-by-step explanation:

you just switch the numbers basically so it would be 3/17