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Natalka [10]
3 years ago
5

For the equation y = -x +5, find y when x = -3

Mathematics
1 answer:
seraphim [82]3 years ago
4 0

Answer:

y= 8

Step-by-step explanation:

y= -x +5

To find the value of y when x= -3, substitute -3 into x and solve the equation.

When x= -3,

y= -(-3) +5

y= 3 +5

y= 8

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Eliminate the parameter and obtain the standard form of the rectangular equation. Ellipse: x = h + a cos(θ), y = k + b sin(θ) Us
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Answer:

a)  (\frac{x-h}{a})^2+ (\frac{y-k}{b})^2=1

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cos^2(\theta)+sin^2(\theta)=1\\(\frac{x-h}{a})^2+ (\frac{y-k}{b})^2=1

this is the equation of an ellipse centered at (h,k), and with horizontal axis length = 2a , and vertical axis length = 2b.

The parametric equations are those we started with:

x=h+a \,cos(\theta)\,,\,y=k+b\,sin(\theta) but we need to find the appropriate parameters for the requested ellipse, as shown below.

For an ellipse of vertices (-4,0) a,d (6,0) and foci at (-2,0) and (4.0), we are dealing with an ellipse with major horizontal axis on the line y=0, and major diameter length of 10 units, so the parameter a=5. The center of the ellipse is therefore at (1,0).

We recall that the vertices of a translated horizontal ellipse are located at (a+h,k) and (-a+h,k), then k=0 to satisfy the information given (-4,0) & (6,0), and since a=5, we deduce that a+h = 6  and therefore h=1.

To find "b" (the only parameter missing for the standard equation of the conic), we need the information on the foci (-2,0) and (4,0) which must equal (h-c,k) and (h+c,k) with k=0 and h=1 which then gives that c=3

Now using the formula for the parameter "c" of the foci: c=\sqrt{a^2-b^2}

c=\sqrt{a^2-b^2}\\3=\sqrt{5^2-b^2}\\9=25-b^2\\b^2=25-9\\b^2=16\\b=4

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