Answer:

Step-by-step explanation:

Lets expand all the composite numbers into prime numbers.

Lets cancel
from numerator and denominator.

Using laws of exponents , lets solve this.


![=> 3^{-3} \times 5^{[1 - (-2)]}](https://tex.z-dn.net/?f=%3D%3E%203%5E%7B-3%7D%20%5Ctimes%205%5E%7B%5B1%20-%20%28-2%29%5D%7D)


Answer:
1
Step-by-step explanation:
1 because we need to round down
Answer:
y = 3x+9
Step-by-step explanation:
We can use the slope intercept form
y = mx+b
where m is the slope and b is the y intercept
y = 3x +b
Substitute the point into the equation
6 = 3(-1) +b
6 = -3+b
Add 3 to each side
6+3 = -3+3+b
9 = b
y = 3x+9
Answer:
(2, 5)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
y - 3x = 1
2y - x = 12
<u>Step 2: Rewrite Systems</u>
y - 3x = 1
- Add 3x on both sides: y = 3x + 1
<u>Step 3: Redefine Systems</u>
y = 3x + 1
2y - x = 12
<u>Step 4: Solve for </u><em><u>x</u></em>
<em>Substitution</em>
- Substitute in <em>y</em>: 2(3x + 1) - x = 12
- Distribute 2: 6x + 2 - x = 12
- Combine like terms: 5x + 2 = 12
- Isolate <em>x</em> term: 5x = 10
- Isolate <em>x</em>: x = 2
<u>Step 5: Solve for </u><em><u>y</u></em>
- Define equation: 2y - x = 12
- Substitute in <em>x</em>: 2y - 2 = 12
- Isolate <em>y </em>term: 2y = 10
- Isolate <em>y</em>: y = 5
Step-by-step explanation:
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