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2 years ago

Is a parallelogram an equilateral?

1 answer:

4 0

"A quadrilateral with both pairs of opposite sides parallel and all sides the same length, i.e., an equilateral parallelogram."

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5x + 7 = 2x - 8 can anyone say how to solve please

SCORPION-xisa [38]

Answer:

x = - 5

Step-by-step explanation:

Given

5x + 7 = 2x - 8 ( subtract 2x from both sides )

3x + 7 = - 8 ( subtract 7 from both sides )

3x = - 15 ( divide both sides by 3 )

x = - 5

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2 years ago

Which congruence criteria is met in the following scenario?

Arisa [49]

None because the triangles have different lines

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2 years ago

A bookstore owner is conducting market research to forecast sales for the coming year. The bookstore is open 360 days a year and

statuscvo [17]

156 dollars because 13 times 12 (months)
equals 156

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2 years ago

You go to a shoe store planning to buy a new pair of the latest Yeezy drop. A pair is selling around for $250. However you have

musickatia [10]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

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