For this case we evaluate the following inequality in the given point and verify if it is met:
Substituting:
It is not fulfilled!
The point does not satisfy the inequality
ANswer:
False
Answer:
For a function y = f(x), the range is the set of all the possible values of y.
In the question you wrote:
y = secx - 2
This can be interpreted as:
y = sec(x - 2)
or
y = sec(x) - 2
So let's see each case (these are kinda the same)
If the function is:
y = sec(x - 2)
Firs remember that:
sec(x) = 1/cos(x)
then we can rewrite:
y = 1/cos(x - 2)
notice that the function cos(x) has the range -1 ≤ y ≤ 1
Then for the two extremes we have:
y = 1/1 = 1
y = 1/-1 = -1
Notice that for:
y = 1/cos(x - 2)
y can never be in the range -1 < x < 1
As the denominator cant be larger, in absolute value, than 1.
Then we can conclude that the range is all reals except the interval:
-1 < y < 1
If instead the function was:
y = sec(x) - 2
y = 1/cos(x) - 2
Then with the same reasoning, the range will be the set of all real values except:
-1 - 2 < y < 1 - 2
-3 < y < -1
B. x<3 is the answer to the question you have asked
The smallest prime number of p for which p^3 + 4p^2 + 4p has exactly 30 positive divisors is 43.
<h3>What is the smallest prime number of p for which p must have exactly 30 positive divisors?</h3>
The smallest number of p in the polynomial equation p^3 + 4p^2 + 4p for which p must have exactly 30 divisors can be determined by factoring the polynomial expression, then equating it to the value of 30.
i.e.
By factorization, we have:
Now, to get exactly 30 divisor.
- (p+2)² requires to give us 15 factors.
Therefore, we can have an equation p + 2 = p₁ × p₂²
where:
- p₁ and p₂ relate to different values of odd prime numbers.
So, for the least values of p + 2, Let us assume that:
p + 2 = 5 × 3²
p + 2 = 5 × 9
p + 2 = 45
p = 45 - 2
p = 43
Therefore, we can conclude that the smallest prime number p such that
p^3 + 4p^2 + 4p has exactly 30 positive divisors is 43.
Learn more about prime numbers here:
brainly.com/question/145452
#SPJ1
Answer:
um yes no maybe and I'll think about it ;)
