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
Answer:
x = 1, y = -2
Step-by-step explanation:
15x + 3y = 9 (1)
10x + 7y = -4 (2)
3 × (2) - 2 × (1)
3 × (2)
30x + 21y = -12 (3)
2 × (1)
30x + 6y = 18 (4)
(3) - (4)
15y = -30
y = -30 / 15
y = -2
sub y = -2 into (1)
15x + 3y = 9
15x + (3 × -2) = 9
15x - 6 = 9
15x = 9 + 6
15x = 15
x = 15 / 15
x = 1
This answer is correct. The absolute value of a number is its distance from 0, and we are looking for an absolute value of 1. Both -1 and 1 are 1 whole number away from 0.
Use the equation of motion under gravity:-
s = 16t^2 where s is the distance and t = time
2063 = 16t^2
t^2 = 2063 / 16 = 128.94 s
t = sqrt 128.94 = 11. 4 seconds to the nearest tenth of a second
Answer:
Its B
Step-by-step explanation:
I did the test on edge
