Answer:
x = 3
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
Step-by-step explanation:
<u>Step 1: Define Equation</u>
17 - (3x + 5) = 3x - 6
<u>Step 2: Solve for </u><em><u>x</u></em>
- Distribute negative: 17 - 3x - 5 = 3x - 6
- Add 3x on both sides: 17 - 5 = 6x - 6
- Combine like terms: 12 = 6x - 6
- Isolate <em>x</em> term: 18 = 6x
- Isolate <em>x</em>: 3 = x
- Rewrite: x = 3
<u>Step 3: Check</u>
<em>Plug in x into the original equation to verify it's a solution.</em>
- Substitute in <em>x</em>: 17 - (3(3) + 5) = 3(3) - 6
- Multiply: 17 - (9 + 5) = 9 - 6
- Add: 17 - 14 = 9 - 6
- Subtract: 3 = 3
Here we see that 3 does indeed equal 3.
∴ x = 3 is the solution to the equation.
Answer:
r = √13
Step-by-step explanation:
Starting with x^2+y^2+6x-2y+3, group like terms, first x terms and then y terms: x^2 + 6x + y^2 -2y = 3. Please note that there has to be an " = " sign in this equation, and that I have taken the liberty of replacing " +3" with " = 3 ."
We need to "complete the square" of x^2 + 6x. I'll just jump in and do it: Take half of the coefficient of the x term and square it; add, and then subtract, this square from x^2 + 6x: x^2 + 6x + 3^2 - 3^2. Then do the same for y^2 - 2y: y^2 - 2y + 1^2 - 1^2.
Now re-write the perfect square x^2 + 6x + 9 by (x + 3)^2. Then we have x^2 + 6x + 9 - 9; also y^2 - 1y + 1 - 1. Making these replacements:
(x + 3)^2 - 9 + (y - 1)^2 -1 = 3. Move the constants -9 and -1 to the other side of the equation: (x + 3)^2 + (y - 1)^2 = 3 + 9 + 1 = 13
Then the original equation now looks like (x + 3)^2 + (y - 1)^2 = 13, and this 13 is the square of the radius, r: r^2 = 13, so that the radius is r = √13.
6x^2-6x+5 would be the correct answer
It would be B because there is more pics of the family than anything else
