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2 years ago

Ellen wrote 18 checks in May. The first five checks had no fee.

1 answer:

6 0

$8.25
The first 5 checks are free so you can subtract the 5 from 18 which is 13. Then you multiply 13 & .25 which is equal to $3.25 and then you just add the $5

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The spinner is divided into a equal sections which two events have the same probability

myrzilka [38]

Answer:

its A

Step-by-step explanation:

7 0

3 years ago

A class has a total of 135 absences over 18 weeks. what is the mean number of absences per week?

Butoxors [25]

135 divided by 18 is 7.5 so I would guess 7 each week and maybe possibly 8 on some weeks

5 0

3 years ago

I don't know how to do this :(

Lemur [1.5K]

Hello.

e is a mathematical constant, and it's like pi, it's irrational. It's approximately equal to 2.71828....

So, $e^{\frac{3}{2}} = 4.4816...$
It's approximately equal to 4.5 which is choice 2.

8 0

3 years ago

Read 2 more answers

Katie is doing an inventory at American Eagle. There is 125 pairs of jeans. 36 pairs of jeans are women's skinny jeans. What per

ZanzabumX [31]

All you have to do is put 36/125. This will get you .288. When rounded to the nearest percent, it is 28%.

3 0

3 years ago

Calculus piecewise function.

Kipish [7]

Part A

The notation $\lim_{x \to 2^{+}}f(x)$ means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.

So we could start at say x = 2.5 and get closer to 2 by getting to x = 2.4 then to x = 2.3 then 2.2, 2.1, 2.01, 2.001, etc

We don't actually arrive at x = 2 itself. We simply move closer and closer.

Since we're on the positive or right hand side of 2, this means we go with the rule involving x > 2

Therefore f(x) = (x/2) + 1

Plug in x = 2 to find that...

f(x) = (x/2) + 1

f(2) = (2/2) + 1

f(2) = 2

This shows $\lim_{x \to 2^{+}}f(x) = 2$

Then for the left hand limit $\lim_{x \to 2^{-}}f(x)$ , we'll involve x < 2 and we go for the first piece. So,

f(x) = 3-x

f(2) = 3-2

f(2) = 1

Therefore, $\lim_{x \to 2^{-}}f(x) = 1$

===============================================================

Part B

Because $\lim_{x \to 2^{+}}f(x) \ne \lim_{x \to 2^{-}}f(x)$ this means that the limit $\lim_{x \to 2}f(x)$ does not exist.

If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.

===============================================================

Part C

You'll follow the same type of steps shown in part A.

However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.

So you'd only involve the second piece f(x) = (x/2) + 1

You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.

===============================================================

Part D

As mentioned earlier, since $\lim_{x \to 4^{+}}f(x) = \lim_{x \to 4^{-}}f(x) = 3$ , this means the limit $\lim_{x \to 4}f(x)$ does exist and it's equal to 3.

As x gets closer and closer to 4, the y values are approaching 3. This applies to both directions.

4 0

1 year ago