Answer:
(1, -2)
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>
4x + y = 2
y = x - 3
<u>Step 2: Solve for </u><em><u>x</u></em>
<em>Substitution</em>
- Substitute in <em>y</em>: 4x + (x - 3) = 2
- Combine like terms: 5x - 3 = 2
- Isolate <em>x</em> term: 5x = 5
- Isolate <em>x</em>: x = 1
<u>Step 3: Solve for </u><em><u>y</u></em>
- Define equation: y = x - 3
- Substitute in <em>x</em>: y = 1 - 3
- Subtract: y = -2
Answer:
$2
Step-by-step explanation:
George borrows $2 so he owes $2.
Answer:
3
Step-by-step explanation:
"" = x^(0.5*(-2))*y^(-0.25*(-2))*z^(-2) = x^(-1)*y^(0.5)*z^(-2) =
(1/x^1)*y^(0.5)*(1/z^2) = 3
Answer:
25.612
Step-by-step explanation:
= 25.612
Answer:
At the end of 6 years, he would have paid $13985.6
Step-by-step explanation:
Initial amount taken as load is $10,000 This means that the principal
P = 10000
It was compounded annually. This means that it was cam pounded once in a year. So
n = 1
The rate at which the principal was compounded is 5.75%. So
r = 5.75/100 = 0.0575
it takes you six years to pay off the loan. So
t = 6
The formula for compound interest is
A = P(1+r/n)^nt
A = total amount of money that you would have paid back by the end of the six years. Therefore
A = 10000 (1+0.0575/1)^1×6
A = 10000(1.0575)^6 = $13985.6
