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igomit [66]
3 years ago
12

1. What is the center and radius of the circle? (x-4)² + (x-7) ² - 49

Mathematics
1 answer:
tester [92]3 years ago
4 0

Answer:

Center at (4, 7) and radius is √49, or 7

Step-by-step explanation:

Didn't you mean (x-4)² + (y-7) ² = 49?

Comparing          (x-4)² + (y-7) ² = 49

to                          (x - h)^2 + (y - k)^2 = r^2, we see that the center is at (h, k)  =>  (4, 7) and that the radius is √49, or 7.

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Combine like terms for <br> (3x²+2x²-5x+7)+(4x²+2x-3)
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Answer:

9x^2-3x+4

Step-by-step explanation:

(3x^2+2x^2-5x+7)+(4x^2+2x-3)

5x^2-5x+7+4x^2+2x-3

5x^2+4x^2-5x+2x+7-3

9x^2-3x+7-3

9x^2-3x+4

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5/x^2+3x-2y is polynomial
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3 years ago
How many solutions does the equation have, x1 + x2 + x3 = 10 , where x1 , x2, and x3 are non-negative integers?
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Non-negative integers are positive integers or zero.

1. When x_1=0, then there are such possible cases for x_2 and x_3:

  • x_2=0,\ x_3=10;
  • x_2=1,\ x_3=9;
  • x_2=2,\ x_3=8;
  • x_2=3,\ x_3=7;
  • x_2=4,\ x_3=6;
  • x_2=5,\ x_3=5;
  • x_2=6,\ x_3=4;
  • x_2=7,\ x_3=3;
  • x_2=8,\ x_3=2;
  • x_2=9,\ x_3=1;
  • x_2=10,\ x_3=0.

In total 11 solutions for x_1=0.

2. For x_1=1, there are such possible cases for x_2 and x_3:

  • x_2=0,\ x_3=9;
  • x_2=1,\ x_3=8;
  • x_2=2,\ x_3=7;
  • x_2=3,\ x_3=6;
  • x_2=4,\ x_3=5;
  • x_2=5,\ x_3=4;
  • x_2=6,\ x_3=3;
  • x_2=7,\ x_3=2;
  • x_2=8,\ x_3=1;
  • x_2=9,\ x_3=0.

In total 10 solutions for x_1=1.

3. This process gives you

  • for x_1=2 - 9 solutions;
  • for x_1=3 - 8 solutions;
  • for x_1=4 - 7 solutions;
  • for x_1=5 - 6 solutions;
  • for x_1=6 - 5 solutions;
  • for x_1=7 - 4 solutions;
  • for x_1=8 - 3 solutions;
  • for x_1=9 - 2 solutions;
  • for x_1=10 - 1 solution.

4. Add all numbers of solutions:

11+10+9+8+7+6+5+4+3+2+1=66.

Answer: there are 66 possible solutions (with non-negative integer variables)

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