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3 years ago

Which of the following ordered pair could NOT be included in this set if it is a function? { (-4, 2), (3, 6), (4, 3), (x, y) }

Mathematics

2 answers:

stellarik [79]3 years ago

7 0

Answer:

C. (x, y) = (3, 6)

Step-by-step explanation: A function is when x's don't repeat.

The correct answer is C. (x, y) = (3, 6) because it would repeating the x's.

Here's Why:

{ (-4, 2), (3, 6), (4, 3), (x, y) }

If you substitute in (3, 6) for the last ordered pair, it would be repeating the number 3. So, (3, 6) can not be included in the set. But the ordered pairs that would NOT be repeating the x's, are these ordered pairs:

A. (x, y) = (1, 2)

B. (x, y) = (-3, 6)

D. (x, y) = (4, -3)

A, B, & D are ordered pairs that can make this set a function.

Hope this helps you!!! :)

sashaice [31]3 years ago

4 0

Answer: (D) 4, -3

Step-by-step explanation:

In order for the set of coordinates to be a function, there cannot be duplicate x-values. So, there cannot be the following: -4, 3, 4

A) x = 1 this CAN be in the set

B) x = -3 this CAN be in the set

C) x = 3 and y = 6 this is a duplicate of a coordinate already in the set. You can plot this point and still pass the vertical line test

D) x = 4 and y = -3 this cannot be included in the set . It will create two different coordinates passing through x = 4 so will fail the vertical line test.

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3 years ago

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3 years ago

Read 2 more answers

The tangent to the circle x^2+y^2=18 is parallel to the tangent x+y=6

liubo4ka [24]

Answer:

y = -x - 6

Step-by-step explanation:

If we are looking for the equation of the line tangent to the circle, we have to find the first derivative of the circle function. If that tangent line is to be parallel to the given tangent line, we need to find the slope of the the given tangent line and make sure the slope of the line we are looking for is the same. First let's find the slope of the given tangent. If the given tangent is x+y=6, then to find the slope, solve for y:

y = -x + 6. So the slope of that line, and also the slope of the tangent we are solving for, is -1. Hold that thought while we find the derivative of the function. Using implicit differentiation, we find the derivative to be:

$2x+2y\frac{dy}{dx}=0$

Solving for the slope (dy/dx) gives us:

$2y\frac{dy}{dx}=-2x$ and

$\frac{dy}{dx}=\frac{-2x}{2y}$ so

$\frac{dy}{dx}=-\frac{x}{y}$

Subbing in the value of the slope we found:

$-1=-\frac{x}{y}$ so

-y = -x or, equivalently,

y = x. Now that we know that y and x are the same value, we can go back to the original circle to find out what they both are by substitution. If y = x, we make the substution:

If $x^2+y^2=18$ , then

$y^2+y^2=18$ and

$2y^2=18$ and

$y^2=9$ so

y = ±3

We will choose the negative root (you'll see why in a second) and sub that in for both y and x since y and x are th same number. Going back to the fact that the slope is -1:

y - (-3) = -1(x - (-3)) simplifies a bit to:

y + 3 = -x - 3 which gives us, in slope-intercept form:

y = -x - 6

That is the equation of the tangent to the circle that is parallel to the given tangent.

If we would have chosen the principle (or positive) root of 3, our equation would have looked like this:

y - 3 = -1(x - 3) and

y = -x + 3 + 3 so

y = -x + 6. Notice that that is the EXACT SAME EQUATION as the given tangent. That's why we have to pick the negative 3 as our root.

Good luck in your calculus class! Make sure you post your questions in here! Many of us LOVE the challenge of calculus!

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