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just olya [345]
3 years ago
7

(a) The parametric equations x = f(t) and y = g(t) give the coordinates of a point (x, y) = (f(t), g(t)) for appropriate values

of t. The variable t is called ___________ .
(b) Suppose that the parametric equations x = t, y = t^2, t >_ 0, model the position of a moving object at time t. When t = 0, the object is at __________ and when t = 6, the object is at ___________.
(c) If we eliminate the parameter in part (b), we get the equation ____________.
Mathematics
1 answer:
Gnesinka [82]3 years ago
5 0

Answer:

Step-by-step explanation:

a). Given a parametric equation, we are describing a set of coordinates based on the value of t. The variable t is called the parameter.

b) we have the following equations. x=t y=t^2, so in order for us to know where the object is at t=t' we must replace t with the specific value t'. Hence, when t=0 the object is at (0,0^2) = (0,0) (the origin). When t=6, the object is at (6,6^2) = (6,36).

c). To eliminate the parameter, we replace the parameter in one equation by using the second equation. Recall that we have that x=t. Then, by replacing in the second equation, we have the following

y=t^2 = (x)^2 = x^2

where x\geq 0

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Answer:y=5x+8

Step-by-step explanation: the perpendicular slope is going to be the negative reciprocal of your given slope. In this case the perpendicular line has a slope of 5, because u flip -1/5 to make -5 and u negate -5 to make 5. This perpendicular slope passes through (-2,-2) and u find the y intercept like this: -2=(5(-2))+b. Therefore the y intercept is 8 and the perpendicular slope is y=5x+8

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What is the equation of a line that passes through the point (6, 1) and is perpendicular to the line whose equation is y=−2x−6y=
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m  =  \frac{1}{2}
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y =  \frac{1}{2} x + b
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passes through the point (6, 1) is
y =  \frac{1}{2} x  - 2
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