Scientific notation question.
1 answer:
Answer: The last option.
Step-by-step explanation:
1. By definition a number written in scientific notation has the following form:
Where <em>a</em> is the coefficient (which is a number between 1 and 10 but not including 10) and <em>b</em> is the exponent (It must be an integer).
2. The decimal point of the coefficient must be wrriten after the first digit.
3. Based on this information, the number shown in the last option has the decimal point after the second digit, therefore, it is not written is scientific notation.
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Given:
A set of composite numbers less than 12.
To find:
The set by listing and set builder methods.
Solution:
Composite number: A positive integer is called a composite number if it has at least one divisor other than 1 and itself.
Composite numbers less than 12 are 4, 6, 8, 9, 10.
By using the listing method , the given set can be expression as:
Set = {4, 6, 8, 9, 10}
The numbers 4, 6, 8, 9, 10 are greater than 1, less than 12 and non prime numbers.
By using set builder method, the given set can be expression as:
Set =
Therefore, the list and set builder notation of the given set are {4, 6, 8, 9, 10} and respectively.
Answer:
maximum: y = 1
minimum: y = 0.
Step-by-step explanation:
Here we have the function:
y = f(x) = √(1 + x^2 - 2x)
we want to find the minimum and maximum in the segment [0, 1]
First, we evaluate in the endpoints, which are 0 and 1.
f(0) =√(1 + 0^2 - 2*0) = 1
f(1) = √(1 + 1^2 - 2*1) = 0
Now let's look at the critical points (the zeros of the first derivate)
To derivate our function, we can use the chain rule:
f(x) = h(g(x))
then
f'(x) = h'(g(x))*g(x)
Here we can define:
h(x) = √x
g(x) = 1 + x^2 - 2x
Then:
f(x) = h(g(x))
f'(x) = 1/2*( 1 + x^2 - 2x)*(2x - 2)
f'(x) = (1 + x^2 - 2x)*(x - 1)
f'(x) = x^3 - 3x^2 + x - 1
this function does not have any zero in the segment [0, 1] (you can look it in the image below)
Thus, the function does not have critical points in the segment.
Then the maximum and minimum are given by the endpoints.
The maximum is 1 (when x = 0)
the minimum is 0 (when x = 1)
