An nth degree<span> polynomial can have as many as n real roots.</span>
Answer:
Continuously
Step-by-step explanation:
Compounded continuously:
A = Pe^(rt)
A = 11,000 e^(0.0625 × 10)
A = 20,550.71
Compounded semiannually (twice per year):
A = P(1 + r)^t
A = 11,000 (1 + 0.063/2)^(2×10)
A = 11,000 (1 + 0.0315)^20
A = 20,453.96
Yeah his answer is correct Good job!
Answer:
True expressions:
- The constants, -3 and -8, are like terms.
- The terms 3 p and p are like terms.
- The terms in the expression are p squared, negative 3, 3 p, negative 8, p, p cubed.
- The expression contains six terms.
- Like terms have the same variables raised to the same powers.
Step-by-step explanation:
The expression is:
p² - 3 + 3p - 8 + p + p³
False expressions:
- The terms p squared, 3 p, p, and p cubed have variables, so they are like terms. (They don't have the same exponents)
- The terms p squared and p cubed are like terms. (They don't have the same exponents)
- The expression contains seven terms. (It contains 6 terms)
