All categories

3 years ago

A) Compare the y-intercepts of f(x) and g(x). Use complete sentences.

Mathematics

2 answers:

Lady bird [3.3K]3 years ago

8 0

Answer:

Step-by-step explanation:

a) y-intercept is when x=0

f(0)

=1/(0-3)

=-1/3

g(0)=0 from the graph

b) vertical asymptotes are vertical lines that f(x) and g(x) get close but do not exist there

Aneli [31]3 years ago

3 0

Answer:

Step-by-step explanation:

a) For the function f, the y-intercept is at (0, -1/3). For the function g as shown in the graph, the y-intercept is at (0,0) and it is higher than f.

b) For the function f, the vertical asymptote is at x=3. For the function g as shown in the graph, the vertical asymptote is at x=4 and is larger than that of f.

You might be interested in

Caroline has 4 green cards, 3 red cars, 3 orange cards, and 1 gold card. if she discards the gold card, what percent of her rema

I am Lyosha [343]

3 reds in the 10 remaining cards
so that is 30% are red

3 0

3 years ago

First right answer gets brilliant. Can someone please help? It is due today!

nikitadnepr [17]

Answer:

Step-by-step explanation:

A, x=31

E, b=11

N, m=61

Y, b=15

I, a=80

G, k=-78

U, x=-13

R, u=88

O, y=13

A, w=104

H, d=-7

V, n=1/36

Sorry, don't want to give away all the answers!

Hope this helps!!!

7 0

3 years ago

) Find the coordinates of the point A' which is the symmetric point

soldier1979 [14.2K]

Let, coordinate of point A' is (x,y).

Since, A' is the symmetric point A(3, 2) with respect to the line 2x + y - 12 = 0.

So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.

Now, their center is given by :

$C( \dfrac{x+3}{2}, \dfrac{y+2}{2})$

Also, product of slope will be -1 .( Since, they are parallel )

$\dfrac{y-2}{x-3} \times -2 = -1\\\\2y - 4 = x - 3\\\\2y - x = 1$

x = 2y - 1

So, $C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})$

Also, C satisfy given line :

$2\times ( \dfrac{2y + 2}{2}) + \dfrac{y+2}{2} = 12\\\\4y + 4 + y + 2 = 24\\\\5y = 18\\\\y = \dfrac{18}{5}$

Also,

$x = 2\times \dfrac{18}{5 } - 1\\\\x = \dfrac{31}{5}$

Therefore, the symmetric points is $A'(\dfrac{31}{5}, \dfrac{18}{5})$ .

7 0

3 years ago

If a=i+7j+k and b=i+11j+k, find a unit vector with positive first coordinate orthogonal to both a and

Scilla [17]

Take the cross product, then normalize the result.

$\mathbf a\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&7&1\\1&11&1\end{vmatrix}=-4\,\mathbf i+4\,\mathbf k$

This has norm

$\|\mathbf a\times\mathbf b\|=\sqrt{(-4)^2+4^2}=4\sqrt2$

and so a unit vector orthogonal to both given vectors is

$\dfrac{\mathbf a\times\mathbf b}{\|\mathbf a\times\mathbf b\|}=-\dfrac1{\sqrt2}\,\mathbf i+\dfrac1{\sqrt2}\,\mathbf k$

An equally correct answer would be the negative of this vector, since $\mathbf b\times\mathbf a=-\mathbf a\times\mathbf b$ .

5 0

3 years ago

Katherine and Shally had $288 altogether. Katherine gave 1/3 of her share to Shally and their father gave $68 to Shally. They no

erastova [34]

1/3 of $288 is $96 So, therefore Shally had $96 at first.
But They both didn't have the same amount of money after too because:
1/3 of $288 is $96
$96 + $68 = $164 So Shally has $164

Now, $288 - $96 = $192. So Katherine has $192

So therefore they did not have the same amount of money but 1/3 of $288 is $96

4 0

3 years ago