No. A polynomial equation in one variablel ooks like P(x) = Q(x), where P and Q are polynomials.
Consider polynomial equations x^2 = 3 and x^2 = 1.
Obviously they have real solutions.
Subtract the two polynomial equations:
(x^2 - x^2) = (3 - 1)
0 = 2...
We get the polynomial equation 0 = 2. We call this a polynomial equation because single constants are also by definition polynomials.
Obviously 0 = 2 has no real solution.
Answer:
f(x) = (x − 3)² − 11
Step-by-step explanation:
To convert to vertex form, complete the square.
f(x) = x² − 6x − 2
f(x) = x² − 6x + 9 − 9 − 2
f(x) = (x − 3)² − 11
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Answer:
Option D
Step-by-step explanation:
Given that the number of hours of daylight in a city in the northern hemisphere shows periodic behavior over time
Let t be the independent variable and no of hours H(t) be the dependent variable on t.
Maximum H(t) = 14.4 and average =12
Period = 365 days
If we fix a sine curve for this since period is 365, we must have coefficient of t as
So the function H(t) will have sine term as sin 0.017t ... i
Since average = 12, we have
H(t) = 12+a sin 0.017t
for some suitable a.
To find a, use maximum
Maximum is when angle is 90 degrees i.e. when sine value =1
max H(5) = 14.4 = 12+a (1)
a =2.4
Hence correct equation would be'H(t) = 2.4sin(0.017t)+12
OPtion D is right answer.