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>
4x + 9 = 2x + 15
<u>Step 2: Solve for </u><em><u>x</u></em>
- Subtract 2x on both sides: 2x + 9 = 15
- Subtract 9 on both sides: 2x = 6
- Divide 2 on both sides: 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>: 4(3) + 9 = 2(3) + 15
- Multiply: 12 + 9 = 6 + 15
- Add: 21 = 21
Here we see that 21 does indeed equal 21.
∴ x = 3 is the solution to the equation.
Answer:
40
Step-by-step explanation:
Ratio of cars to total:
32:80
so we can make another ratio with x
x:100
now make them fractions and cross-multiply
32/80 = x/100
32(100) = 80x
80x = 3200
x = 40
The answer is 7. So you would do
2(3)2=12
4(3)=12
12-12=0
0+7
7
Step-by-step explanation:
The are of rectangle is 1300.5cm2
Answer: (x-2)^2+(y+3)^2 = 9Side notes
1) This circle has a center of (2,-3)
2) The radius of this circle is 3
3) The graph is shown in the attached image
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Work Shown:
x^2-4x+y^2+6y+4=0
x^2-4x+y^2+6y+4-4=0-4
x^2-4x+y^2+6y = -4
x^2-4x+4+y^2+6y = -4+4 ... see note 1 below
(x^2-4x+4)+y^2+6y = 0
(x-2)^2+y^2+6y = 0
(x-2)^2+y^2+6y+9 = 0+9 ... see note 2 below
(x-2)^2+(y^2+6y+9) = 9
(x-2)^2+(y+3)^2 = 9note 1: I'm adding 4 to both sides to complete the square for the x terms. You do this by first taking half of the x (not x^2) coefficient which in this case is -4. So take half of -4 to get -2. Then square this result to get 4
note 2: Like with note 1, I'm completing the square. What's different this time is that this is for the y terms now. The y coefficient is 6. Half of this is 3. Square 3 to get 9. So this is why we add 9 to both sides.
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So
the equation in standard form is (x-2)^2+(y+3)^2 = 9Note how
(x-2)^2+(y+3)^2 = 9
is equivalent to
(x-2)^2+(y-(-3))^2 = 3^2
So that second equation listed above is in the form (x-h)^2+(y-k)^2 = r^2
where
h = 2
k = -3
r = 3
making the center to be (h,k) = (2,-3) and the radius to be r = 3
The graph is attached.