Combine like terms:
n^3 + 20n^ + 10n^2
= n^3 + 30n^
Divide:
=n^3 + 30n^2/10n^2
Factor:
=n^2n + 30n^2
=n^(n + 30)
= 2^2(n + 30)/10n^2
Cancel the common factor n^2:
=n + 10/10
<u>(b+h)(x)=</u><u>16x+2</u>
Since we all know the theory that shows that a(x)+b(x)=(a+b)(x), we can use that in this problem. Simply simplify the problem and divide it into 2... b(x) plus h(x). This is simple as we already know the values of these variables so just add both and the sum will be the answer to your problem.
Answer:
(-3, -15)
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>
n = 5m
n = 2/3m - 13
<u>Step 2: Solve for </u><em><u>m</u></em>
<em>Substitution</em>
- Substitute in <em>n</em>: 5m = 2/3m - 13
- Subtract 2/3m on both sides: 13/3m = -13
- Divide 11/3 on both sides: m = -3
<u>Step 3: Solve for </u><em><u>n</u></em>
- Define equation: n = 5m
- Substitute in <em>m</em>: n = 5(-3)
- Multiply: n = -15
Answer:
$1.20 per chocolate bar
Step-by-step explanation:
$18/15 choco bars
xxx=$18.00/15 bars
$18.00/15/bars
$1.20/bars
y = |x - 3|
3x + 3y = 27
3x + 3|x - 3| = 27
3x + 3|x - 3| = ±27
3x + 3|x - 3| = 27 or 3x + 3|x - 3| = -27
3x + 3(x - 3) = 27 or 3x + 3(x - 3) = -27
3x + 3(x) - 3(3) = 27 or 3x + 3(x) - 3(3) = -27
3x + 3x - 9 = 27 or 3x + 3x - 9 = -27
6x - 9 = 27 or 6x - 9 = -27
+ 9 + 9 + 9 + 9
6x = 36 or 6x = -18
6 6 6 6
x = 6 or x = -3
y = |x - 3| or y = |x - 3|
y = |6 - 3| or y = |-3 - 3|
y = |3| or y = |-6|
y = 3 or y = 6
(x, y) = (6, 3) or (x, y) = (-3, 6)
The two systems of equations of the graph is only equal to (6, 3). It is not equal to (-3, 6) because one of the systems of equations - y = |x - 3| - only has one solution to the function. So the answer to the problem is 3 - (6, 3) is the solution to the system because it satisfies both equations.