Answer:
(-2, 20)
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 = -7x + 6
y = -10x
<u>Step 2: Solve for </u><em><u>x</u></em>
<em>Substitution</em>
- Substitute in <em>y</em>: -10x = -7x + 6
- Add 10x to both sides: 0 = 3x + 6
- Isolate <em>x</em> term: -6 = 3x
- Isolate <em>x</em>: -2 = x
- Rewrite: x = -2
<u>Step 3: Solve for </u><em><u>y</u></em>
- Define equation: y = -10x
- Substitute in <em>x</em>: y = -10(-2)
- Multiply: y = 20
Answer:
Yes the integral can be evaluated by integration by parts as solved below.
Step-by-step explanation:
Taking algebraic function as first function and exponential function as second function we have
Answer:
a
Step-by-step explanation:
Question: solve for r: 6 ^ (2r) = 800
Answer: 1.87
Step-by-step explanation:
6 ^ (2r) = 800
log(6 ^ (2r)) = log(800)
2r log(6) = log(800)
r = log(800) / (2 log(6))
r = 1.86537641979
round
r = 1.87
log can be any logarithm function.
6^(2×1.86) = 784.734328546
6^(2×1.87) = 813.365367336
