Which represents the domain of the following relation? {(-6, 5), (-4, 3), (-1, 0), (4, 3)}

What is 1 5/6 + 1/2 ? a. 14/6 b. 17/12 c. 17/6 d. 25/12

Vladimir79 [104]

Make bottom numbers same

convert to improper fraction

1 and 5/6=1+5/6=6/6+5/6=11/6

1/2 and 11/6
2 times 3=6
1/2 times 3/3=3/6

1/2+11/6=3/6+11/6=(4+11)/6=14/6

answer is A

3 0

3 years ago

Greg purchases 1/2 dozens cupcakes for 5.88. Explain how you can determine the unit cost of the cupcakes.

Reika [66]

Answer: $0.98

Step-by-step explanation:

1/2 dozen is equal to 6 cupcakes

Divide 5.88 by 6 to get 0.98

This is the price for 1 cupcake, which is also the unit price.

7 0

3 years ago

The equation below represents Function A and the graph represents Function B:

kozerog [31]

The slope of function A is the coefficient of x, which is 6.

The slope of function B is the rise divided by the run, which is 3/1 = 3.

The appropriate choice is ...

... c) Slope of Function A = 2×Slope of Function B

8 0

3 years ago

What is the landing strip’s area ?

Vinil7 [7]

The answer is D.
First subtract 9 from 5,364 then divide you answer by two. Then you should get 2673. Multiple this number by 9 to get an area of 24,057.

5 0

2 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