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3 years ago

If a graph passes the horizontal line test, but not the vertical line test, is it still a function?

Mathematics

1 answer:

nydimaria [60]3 years ago

8 0

If a graph fails the vertical line test it's not a function; you can't have a single x map to two ys.

Vice versa, if the graph passes the verticle line test, it's a function.

The horizontal line test tests for injectivity aka one-to-one-ness; it has no bearing on whether something's a function.

You might be interested in

Please help and please say the formula!!

Damm [24]

Answer:

Step-by-step explanation:

Formula: 4*((15*7)/2) + 15^2

This is the area of the four triangles combined (210 cm^2) plus the area of the square (225 cm^2)

8 0

3 years ago

There are 20 players on a soccer team. From them, a captain and an alternate captain have to be chosen. How many possibilities a

tester [92]

Answer:

190 possibilities.

Step-by-step explanation:

The problem is called 20 choose 2, given by the formula for combinations

C(20,2) = 20! / (2! (20-2)!) = 190

alternatively, we can reason it this way:

there are 20 choices to choose a captain and 19 choices for an alternate captain for a total of 380 ways.

However, there are 20 choices to choose an alternate captain and 19 choices for a captain for a total of 380 ways.

Since we are essentially double counting every choice of the posts, we need to divide 380 by 2 to get 190 ways.

3 0

3 years ago

Circle P is described by the equation (x−1)2+(y+6)2=9 and circle Q is described by the equation (x+4)2+(y+14)2=4. Select from th

xeze [42]

Answer:

(a) Circle Q is 9.4 units to the center of circle P

(b) Circle Q has a smaller radius

Step-by-step explanation:

Given

$P:(x - 1)^2 + (y + 6)^2 = 9$

$Q:(x + 4)^2 + (y + 14)^2 = 4$

Solving (a): The distance between both

The equation of a circle is:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$Center: (h,k)$

$Radius:r$

P and Q can be rewritten as:

$P:(x - 1)^2 + (y + 6)^2 = 3^2$

$Q:(x + 4)^2 + (y + 14)^2 = 2^2$

So, for P:

$Center = (1,-6)$

$r = 3$

For Q:

$Center = (-4,-14)$

$r = 2$

The distance between them is:

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

Where:

$Center = (1,-6)$ --- $(x_1,y_1)$

$Center = (-4,-14)$ --- $(x_2,y_2)$

So:

$d = \sqrt{(1 - -4)^2 + (-6 - -14)^2$

$d = \sqrt{(5)^2 + (8)^2$

$d = \sqrt{25 + 64$

$d = \sqrt{89$

$d = 9.4$

Solving (b): The radius;

In (a), we have:

$r = 3$ --- circle P

$r = 2$ --- circle Q

By comparison

$2 < 3$

Hence, circle Q has a smaller radius

4 0

3 years ago

Find the directional derivative of the function at the given point in the direction of the vector v. G(r, s) = tan−1(rs), (1, 3)

alexandr1967 [171]

The directional derivative of $f$ at the given point in the direction indicated is $\frac{5}{2}$ .

<h3>How to calculate the directional derivative of a multivariate function</h3>

The directional derivative is represented by the following formula:

$\nabla_{\vec v} f = \nabla f (r_{o}, s_{o})\cdot \vec v$ (1)

Where:

$\nabla f (r_{o}, s_{o})$ - Gradient evaluated at the point $(r_{o}, s_{o})$ .
$\vec v$ - Directional vector.

The gradient of $f$ is calculated below:

$\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{\partial f}{\partial r}(r_{o},s_{o}) \\\frac{\partial f}{\partial s}(r_{o},s_{o}) \end{array}\right]$ (2)

Where $\frac{\partial f}{\partial r}$ and $\frac{\partial f}{\partial s}$ are the partial derivatives with respect to $r$ and $s$ , respectively.

If we know that $(r_{o}, s_{o}) = (1, 3)$ , then the gradient is:

$\nabla f(r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{s}{1+r^{2}\cdot s^{2}} \\\frac{r}{1+r^{2}\cdot s^{2}}\end{array}\right]$

$\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{3}{1+1^{2}\cdot 3^{2}} \\\frac{1}{1+1^{2}\cdot 3^{2}} \end{array}\right]$

$\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{3}{10} \\\frac{1}{10} \end{array}\right]$

If we know that $\vec v = 5\,\hat{i} + 10\,\hat{j}$ , then the directional derivative is:

$\nabla_{\vec v} f = \left[\begin{array}{cc}\frac{3}{10} \\\frac{1}{10} \end{array}\right] \cdot \left[\begin{array}{cc}5\\10\end{array}\right]$

$\nabla _{\vec v} f (r_{o}, s_{o}) = \frac{5}{2}$

The directional derivative of $f$ at the given point in the direction indicated is $\frac{5}{2}$ . $\blacksquare$

To learn more on directional derivative, we kindly invite to check this verified question: brainly.com/question/9964491

3 0

2 years ago

PLEASE HELP ITS URGENT I HAVE 3 MINS AND IM FAILING I REALLY NEED A A ASAP PLEASE HELP

Igoryamba

Answer:

I am pretty sure it is B I am not sure but this is what I think

Step-by-step explanation:

5 0

3 years ago