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sergij07 [2.7K]
3 years ago
5

How do you reflect over y axis

Mathematics
2 answers:
Katena32 [7]3 years ago
5 0
Think of a point not on the y-axis.
For example, think of the point (5, 5).
This point is in quadrant 1, right of the y-axis and above the x-axis.
To reflect it over the y-axis, the reflection will be in quadrant 2, still above the x-axis, but now left of the y-axis. The distance the reflection is from the y-axis is the same as the original point is from the y-axis.

The reflection of (5, 5) over the y-axis is (-5, 5).

To reflect any point over the y-axis, replace the x-coordinate with the additive inverse of the x-coordinate.
aleksley [76]3 years ago
3 0
If you want to reflect over the y axis then your answer would want to be going either from the left side to the right side, or right side over to the left side. If the figure was on the top or bottom you would go over the x axis.

So to reflect over the y axis you must find the points that each angle or corner of the figure your reflecting. You'll want to remember those locations of the points, then you'll transfer them over to the other side on the same points but on the other side. I'll make an example to make it more clear

For the first number, before the coma, if the number is negative you'll go left, if the number is positive you go right
For the second number, after the coma, you go up if it's positive or down if its negative.

If A is located at (-3, +2)
Angle B at (-1, +2)
Angle C at (-2, +0)

To transfer them over the y axis the numbers being positive or negative will change
For example

Angle A will become (+3, +2)
Angle B will become (+1, +2)
And angle C will become (+2, 0)

If the number is zero then it stays on the default line.

Hope this helped,reply if you need more help!!
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A company manufactures running shoes and basketball shoes. The total revenue (in thousands of dollars) from x1 units of running
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Answer:

x_1 =2 , x_2=7

Step-by-step explanation:

Consider the revenue function given by R(x_1,x_2) = -5x_1^2-8x_2^2 -2x_1x_2+34x_1+116x_2. We want to find the values of each of the variables such that the gradient( i.e the first partial derivatives of the function) is 0. Then, we have the following (the explicit calculations of both derivatives are omitted).

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From the first equation, we get, x_2 = \frac{-10x_1+34}{2}.If we replace that in the second equation, we get

-16\frac{-10x_1+34}{2} -2x_1+116=0= 80x_1-2x_1+116-272= 78x_1-156

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x_2 = \frac{-10\cdot 2 +34}{2}=\frac{14}{2} = 7

So, the critical point is (x_1,x_2) = (2,7). We must check that it is a maximum. To do so, we will use the Hessian criteria. To do so, we must calculate the second derivatives and the crossed derivatives  and check if the criteria is fulfilled in order for it to be a maximum. We get that

\frac{d^2R}{dx_1dx_2}= -2 = \frac{d^2R}{dx_2dx_1}

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We have the following matrix,  

\left[\begin{matrix} -10 & -2 \\ -2 & -16\end{matrix}\right].

Recall that the Hessian criteria says that, for the point to be a maximum, the determinant of the whole matrix should be positive and the element of the matrix that is in the upper left corner should be negative. Note that the determinant of the matrix is (-10)\cdot (-16) - (-2)(-2) = 156>0 and that -10<0. Hence, the criteria is fulfilled and the critical point is a maximum

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3 years ago
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