All categories

3 years ago

Which of the following is a result of shifting a circle with equation (x + 3)² + (y - 2)² = 36 left 3 units?

2 answers:

5 0

A circle with equation:
( x + 3 )² + ( y - 2 )² = 36
Center of a circle: C ( - 3, 2 )
r = 6 ( the radius of a circle stays the same )
After shifting 3 units left : C 1 ( - 6, 2 ).
Answer:
( x + 6 )² + ( y - 2 )² = 36

Neporo4naja [7]3 years ago

3 0

Answer:the y-coordinate of the center of the circle increases by 3

Step-by-step explanation:

You might be interested in

Which graph represents the function?

mario62 [17]

For this function we can find y-intercept.
x=0, y=-2
This graph is on the top, right.

3 0

3 years ago

You are buying boxes of cookies at a bakery. Each box of cookies costs 4 dollars. In the equation below, c represents the number

Yuliya22 [10]

What the answer you want to find out ?

7 0

3 years ago

The points (-4, 3) and (4, r) lie on a line with slope 1/2<br> Find the missing coordinate r.

melisa1 [442]

Answer:

Step-by-step explanation:

8 units separate -4 and 4

8 divided by 2 (which is the x of slope) is four

Four times 1 (which is y of slope) equals four

3+4 equals seven

(4,7)

7 0

3 years ago

Help me out please :(?

Andre45 [30]

Answer:

3x2+15x=18

Step 1: Subtract 18 from both sides.

3x2+15x−18=18−18

3x2+15x−18=0

Step 2: Factor left side of equation.

3(x−1)(x+6)=0

Step 3: Set factors equal to 0.

x−1=0 or x+6=0

x=1 or x=−6

7 0

3 years ago

Read 2 more answers

F(x)=x-1/x^2-x-6 which is the graph of

hoa [83]

Answer:

The graph is attached below.

Step-by-step explanation:

<em>As you have not added the graph, so I will be solving the function for a graph.</em>

Given the function

$f\left(x\right)=x-\frac{1}{x^2}-x-6$

$x-\mathrm{axis\:interception\:points\:of\:}-\frac{1}{x^2}-6:$

$\mathrm{x-intercept\:is\:a\:point\:on\:the\:graph\:where\:}y=0$

$-\frac{1}{x^2}-6=0$

$-1-6x^2=0$

$\mathrm{No\:Solution\:for}\:x\in \mathbb{R}$

$\mathrm{No\:x-axis\:interception\:points}$

$y-\mathrm{axis\:interception\:point\:of\:}-\frac{1}{x^2}-6:$

$y\mathrm{-intercept\:is\:the\:point\:on\:the\:graph\:where\:}x=0$

As we know that the domain of a function is the set of input or argument values for which the function is real and defined.

$\mathrm{Domain\:of\:}\:-\frac{1}{x^2}-6\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x0\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:0\right)\cup \left(0,\:\infty \:\right)\end{bmatrix}$

$\mathrm{Since}\:x=0\:\mathrm{is\:not\:in\:domain}$

$\mathrm{No\:y-axis\:interception\:point}$

$\mathrm{Asymptotes\:of}\:-\frac{1}{x^2}-6:\quad \mathrm{Vertical}:\:x=0,\:\mathrm{Horizontal}:\:y=-6$

$\mathrm{Range\:of\:}-\frac{1}{x^2}-6:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)$

The graph is attached below.

5 0

3 years ago