Answer:
15f^2
Step-by-step explanation:
multiply lenghtxwidth
always make sure to square your answer when doing area
So perimeter is equal to 2 lengths and 2 width so the equation would be P=2L+2W. you know width so you plug in 5 for W then solve for L and you should get L=3
Answer:
(8, -8)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality
<u>Algebra I</u>
- Terms/Coefficients
- Coordinates (x, y)
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
y = x - 16
5y = 2x - 56
<u>Step 2: Solve for </u><em><u>x</u></em>
- Substitute in <em>y</em>: 5(x - 16) = 2x - 56
- Distribute 5: 5x - 80 = 2x - 56
- [Subtraction Property of Equality] Subtract 2x on both sides: 3x - 80 = -56
- [Addition Property of Equality] Add 80 on both sides: 3x = 24
- [Division Property of Equality] Divide 3 on both sides: x = 8
<u>Step 3: Solve for </u><em><u>y</u></em>
- Define original equation: y = x - 16
- Substitute in <em>x</em>: y = 8 - 16
- Subtract: y = -8
Answer:
Shaneeka live from Phil 7/33 mile far.
Step-by-step explanation:
Given:
Ben lives 6/11 mile north of Phil, and 1/3 mile north of Shaneeka.
Now, to get how far does Shaneeka live from Phil.
So, by subtracting the distance from 6/11 mile north where Ben lives from Phil to 1/3 mile north where Ben lives from Shaneeka:
Therefore, Shaneeka live from Phil 7/33 mile far.
