Please help find domain and range
2 answers:
<h3>Answer: Choice A</h3>
- Domain: x > 4
- Range: y > 0
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Explanation:
We want to avoid having a negative number under the square root. Solving leads to
So it appears the domain could involve x = 4 itself; however, if we tried that x value, then we'd get a division by zero error.
So in reality, the domain is x > 4.
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The range of y = sqrt(x) is the set of positive real numbers. So y > 0 is the range for this equation. Shifting left and right does not affect the range, so the range of y = sqrt(x-4) is also y > 0.
We are dividing a positive number (3) over some positive number in the denominator. Overall, the expression is positive because positive/positive = positive.
Therefore, the range of the given equation is y > 0
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The graph is shown below. We have a vertical asymptote at x = 4 and a horizontal asymptote at y = 0. The green curve is fenced in the upper right corner (northeast corner).
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Answer:
- a)
- b)
- c)
- d)
Step-by-step explanation:
We will use the product rule from combinatorics.
- a) There are 26 letters in the English alphabet, so there are 26 possible choices for the first character and 26 possible choices for the last one. Each one of the remaining eight characters of the string has 36 choices (letters or digits). By the product rule, there are
strings.
- b) We have 5 possible choices for the first character, it must be some vowel a,e,i,o,u. The second character can be chosen in 21 ways, selecting some consonant. There are 10 possibilities for the last character because only of the digits are allowed. The other seven characters have no restrictions, so each one can be chosen in 36 ways. By the product rule there are
strings.
- c) The third character has 5 possibilities. Repetition of vowels is allowed, so the sixth and eighth characters have each one 5 possible choices. There are seven characters left. None of them are a vowel, but they are allowed to take any other letter or digit, so each one of them can be chosen in 36-5=31 ways. Therefore there are
strings.
- d) Remember that the binomial coefficient
is the number of ways of choosing k elements from a set of n elements. In this case, to count all the possible strings, we first need to count in how many ways we can select the four positions that will have the digits. This can be done in
ways, since we are choosing four elements from the set of the ten positions of the string. Now, for the first position, we can choose any digit so it has 10 possibilities. The second position has 9 possibilities, because we can't repeat the digit used on the first position. Similarly, there are 8 choices for the third position and there are 7 choices for the fourth. Now, these are the only digits on the string, so the remaining 6 characters must be letters, then each one of them has 26 possibilities. By the product rule, there are
strings.