Can someone help me figure out the answer to:3 5/8 + 4 2/3?(please explain working?)

1 answer:

3 0

You would first make the fractions improper (making the numerator bigger then the denominator) in order to do that times the denominator with the whole number and add the numerator.
Ex) 3*8 = 24 + 5 = 29/8
Then you would add both of the improper fractions

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PLEASE HELP!! A classmate wrote the set notation for the values of the domain of a function {−5, −4, −3, −2, −1, 0} as {x | −5 ≤

vfiekz [6]

<h3>Answer: Choice B</h3>

The set notation includes all values from -5 to 0, but the domain only includes the integer values

eg: something like -1.2 is in the second set, but it is not in the set {-5,-4,-3,-2,-1}

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Further explanation:

Let's go through the answer choices one by one

A. This is false because 0 does not come before -5, but instead -5 is listed first. The order -5,-4,-3,-2,-1,0 is correct meaning that $-5 \le x \le 0$ is the correct order as well.
B. This is true. A value like x = -1.2 is in the set $\left\{x| -5 \le x \le 0\right\}$ since -1.2 is between -5 and 0; but -1.2 is not in the set {-5, -4, -3, -2, -1, 0}. So the distinction is that we're either considering integers only or all real numbers in this interval. To ensure that we only look at integers, the student would have to write $\left\{x| x\in\mathbb{Z}, \ -5 \le x \le 0\right\}$ . The portion $x \in \mathbb{Z}$ means "x is in the set of integers". The Z refers to the German word Zahlen, which translates to "numbers".
C. This is false. The student used the correct inequality signs to indicate x is -5 or larger and also 0 or smaller; basically x is between -5 and 0 inclusive of both endpoints. The "or equal to" portions indicate we are keeping the endpoints and not excluding them.
D. This is false. Writing $-5 \ge x \ge 0$ would not make any sense. This is because that compound inequality breaks down into $-5 \ge x \ \text{ and } \ x \ge 0$ . Try to think of a number that is both smaller than -5 AND also larger than 0. It can't be done. No such number exists.