Ask question

Ask question

All categories

3 years ago

14

On a coordinate plane, a circle has a center at point (negative 6, 4). What is the equation of the circle shown in the graph? (

)² + ()² =

2 answers:

lukranit [14]3 years ago

5 0

Answer:

1) x+6 2)y-4 3)36

Step-by-step explanation:

(x+6)² + (y-4)² = 36

Archy [21]3 years ago

4 0

Answer:

First Drop down: X+6

Second: Y-4

Third: 36

Step-by-step explanation:

You might be interested in

How many unit cubes are needed to make the large cube?

DaniilM [7]

Answer:

D. 64 should be the answer

Step-by-step explanation:

length x width x height

4 x 4 x 4

4^2

64

7 0

2 years ago

Combine terms to simplify a + 2a +7

Nimfa-mama [501]

Answer:

Step-by-step explanation:

Identify Like Terms: <em>a</em> and <em>2a</em> are like
Combine Like Terms: <em>a + 2a = 3a</em>
Substitute to Form an Expression: <em>3a + 7</em>

Sincerely,

<em>Gigabyte</em>

8 0

2 years ago

Help I need to submit soon

Ostrovityanka [42]

Answer:

4

Step-by-step explanation:

I don't know if u want the explanation

8 0

2 years ago

Read 2 more answers

What is the total surface area of the drawing?

kkurt [141]

The answer will be 21.2kilometers

7 0

3 years ago

Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$ 's equal to find those $x$ 's for which the $y$ 's are the same.

$\log(x)=\frac{1}{2}\log(x+1)$

By power rule:

$\log(x)=\log((x+1)^\frac{1}{2})$

Since $\log(u)=\log(v)$ implies $u=v$ :

$x=(x+1)^\frac{1}{2}$

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

$b=-1$

$c=-1$

Let's plug them into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

$x=\frac{1+\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}$ .

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$ .

Let's find the $y$ -coordinate.

You may use either equation.

I choose $y=\log(x)$ .

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

6 0

2 years ago