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Alex Ar [27]
3 years ago
8

The circle below is centered at (3 1) and has a radius of 2 what is its equation

Mathematics
1 answer:
Dominik [7]3 years ago
7 0

Answer:

(x-3)^2 + (y-1)^2 = 4

Step-by-step explanation:

The equation of a circle is usually written in the form

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

Where (h,k) is the center and r is the radius

The center is at the origin so (h,k) = (3,1) and  the radius is 2 so r=2

(x-3)^2 + (y-1)^2 = 2^2

(x-3)^2 + (y-1)^2 = 4

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Mekhanik [1.2K]

Given:

The function is:

f(x)=x^2-x+1

To find:

The result of the operation -f(x)=-(x^2-x+1).

Solution:

If g(x)=-f(x), then the graph of f(x) is reflected across the x-axis to get the graph of g(x).

We have,

f(x)=x^2-x+1

The given operation is:

-f(x)=-(x^2-x+1)

So, it will result in a reflection across the x-axis.

Therefore, the correct option is A.

4 0
2 years ago
How do you put 943,261,586 with base ten numbers
Citrus2011 [14]

Answer:

(9 \times 10^{8}) + (4 \times 10^{7}) + (3 \times 10^{6}) + (2 \times 10^{5}) +  (6 \times 10^{4}) +  (10^{3}) +  (5 \times 10^{2}) +  (8 \times 10) + (6 \times 10^{0})

Step-by-step explanation:

How do you put 943,261,586 with base ten numbers

943,261,586  =  

900,000,000 + 40,000,000 + 3,000,000 + 200,000 + 60,000 + 1,000 + 500 + 80 +6 =

(9 \times 10^{8}) + (4 \times 10^{7}) + (3 \times 10^{6}) + (2 \times 10^{5}) +  (6 \times 10^{4}) +  (10^{3}) +  (5 \times 10^{2}) +  (8 \times 10) + (6 \times 10^{0})

8 0
2 years ago
If (2x - 3) is a factor of mx^2 - 11x - 6, find the other factor.​
Travka [436]

Answer:

(2x - 3) ^2 - 11x - 6

Step-by-step explanation:

6 0
2 years ago
How to find imaginary zeros anf real zeros of F(x)=-4x^5-8x^3+12x​
Fofino [41]

Answer:

x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}

Step-by-step explanation:

We are given the function:

f(x)=-4x^5-8x^3+12x

And we want to finds its zeros.

Therefore:

0=-4x^5-8x^3+12x

Firstly, we can divide everything by -4:

0=x^5+2x^3-3x

Factor out an x:

0=x(x^4+2x^2-3)

This is in quadratic form. For simplicity, we can let:

u=x^2

Then by substitution:

0=x(u^2+2u-3)

Factor:

0=x(u+3)(u-1)

Substitute back:

0=x(x^2+3)(x^2-1)

By the Zero Product Property:

x=0\text{ and } x^2+3=0\text{ and } x^2-1=0

Solving for each case:

x=0\text{ and } x=\pm\sqrt{-3}\text{ and } x=\pm\sqrt{1}

Therefore, our real and complex zeros are:

x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}

4 0
2 years ago
Help plz i have a f in this class
Mnenie [13.5K]

Answer:

For number 3 the answer is -4

4 0
2 years ago
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