sold 45 of his DVDs and purchased 10 more. If Smith has 15 DVDs now, how many DVDs did he have to begin with?
1 answer:
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Answer:
For the perfect square trinomial (quadratic) i.e. , the constant term (last term) is positive.
Step-by-step explanation:
"Perfect square trinomials" are termed as the quadratics that are the outcomes of squaring binomials.
For example:
Therefore, for the perfect square trinomial (quadratic) i.e. , the constant term (last term) is positive.
Answer:
N(1)=50 is a minimum
N(15)=4391.7 is a maximum
Step-by-step explanation:
<u>Extrema values of functions </u>
If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.
We are given a function (corrected)
(a)
First, we take its derivative
Solve N'(t)=0
Simplifying
Solving for t
Only t=1 belongs to the valid interval
Taking the second derivative
Which is always positive, so t=1 is a minimum
(b)
N(1)=50 is a minimum
(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15
(d)
N(15)=4391.7 is a maximum
Answer:
B, E, and F.
Step-by-step explanation:
We can apply synthetic division when dividing a polynomial P(x) by a binomial in the form (x - a).
Looking at the coefficients, we can see that our dividend is:
Our divisor will be (x - a), where a is the number on the very top left.
So, a = 2.
Thus, our divisor is:
Note that our remainder is 0.
So, the synthetic division shown is telling us that:
Hence, E is the correct choice. A uses the incorrect divisor.
If the divisor successfully divides with no remainder, then the divisor is a factor.
Since (x - 2) successfully divided with no remainder, (x - 2) is indeed a factor of the polynomial.
Hence, B is also correct.
C uses the wrong factor.
D uses the wrong value. It should be positive 2.
So, F is the correct choice.
Therefore, the answers are: B, E, and F.