Answer:
Infinite amount of solutions
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>
</u>
<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 = -2x + 4
2x + y = 4
<u>Step 2: Solve for </u><em><u>x</u></em>
<em>Substitution</em>
- Substitute in <em>y</em>: 2x + (-2x + 4) = 4
- Combine like terms: 4 = 4
Here we see that 4 does indeed equal 4.
∴ the systems of equations has an infinite amount of solutions.
The area of a regular hexagon is
... A = (3√3)/2×s² . . . . . where s is the side length
Of course, the 6-sided figure will have a side length that is 1/6 of the perimeter.
... s = (48 in)/6 = 8 in
... A = (3√3)/2×(8 in)² = 96√3 in² . . . area of your regular hexagon
A. The discount is 7.80 much
B. $11.7
Answer:
a = 9
Step-by-step explanation:
First we need to add (x^2+3x) to (3x^2+ax),
(x^2+3x)+(3x^2+ax)
Expand
= x^2+3x+3x^2+ax
Collect the like terms
= x^2+3x^2+3x+ax
= 4x^2 + 3x + ax
= 4x^2+(3+a)x
Equate the solution to 4x^2+12x
4x^2+(3+a)x = 4x^2+12x
Comparing the like terms on both sides
(3+a)x =12x
3 + a = 12
a = 12 - 3
a = 9
Hence the value of a is 9
The volume is (area of cross-section) x (length) .
-- The cross-section is a triangle. The area of a triangle is
Area = (1/2) (base) (height) .
In this one, the base is 9/4 m and the height is 3-1/3 m .
Area = (1/2) (9/4 m) (3-1/3 m)
Area = (1/2) (9/4) (10/3)
Area = 90/24 m² .
-- Volume = (area of cross-section) x (length)
Volume = (90/24 m²) x (7-1/3 m)
Volume = (90/24) x (22/3) m³
Volume = (1,980 / 72) m³
<em>Volume = 27.5 m³ </em>
