Answer:
(p + q)² - ∛(h·3k) or (p + q)² - ∛(h·3k)
Step-by-step explanation:
Cube root of x: ∛x
Product of h and 3k: h·3k
Sum of p and q: p + q
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From (p + q)² subtract ∛(h·3k) This becomes, symbolically:
=> (p + q)² - ∛(h·3k)
Answer:
1
Step-by-step explanation:
to lazy
Answer:
1. Technically 2, but might be 0 in your teacher's opinion.
2. 1
Step-by-step explanation:
Solving problem one.
So I don't know if you have learned about imaginary numbers, but if you have, then you would end up with two answers if you plugged in the quadratic formula.
If you haven't learned about imaginary numbers, then I would say your best option would be to write 'No real solution' since there are technically 2 solutions.
Solving problem two.
Turns out this quadratic has a special property and it's actually a square of one equation. You can find out by just factoring the equation.
It's (3x-2)^2. Since it's squared, that means that only 2/3 would work as x in this equation.
Answer:
g(f(2)) = 11
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
Step-by-step explanation:
<u>Step 1: Define</u>
f(x) = 3x - 2
g(x) = x² - 5
<u>Step 2: Find g(f(2))</u>
f(2)
- Substitute in <em>x</em>: f(2) = 3(2) - 2
- Multiply: f(2) = 6 - 2
- Subtract: f(2) = 4
g(f(2))
- Substitute in <em>x</em>: g(4) = 4² - 5
- Exponents: g(4) = 16 - 5
- Subtract: g(4) = 11
