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3 years ago

45+5/6x= 50 Help me find the answer please

Mathematics

1 answer:

belka [17]3 years ago

6 0

Answer:

Step-by-step explanation:

45 + 5/6x = 50.....subtract 45 from both sides

5/6x = 50 - 45

5/6x = 5....now multiply both sides by 6/5 to get rid of the 5/6 on the left

(6/5)(5/6)x = 5(6/5)

x = 30/5 which reduces to 6 <==

45 + 5/6x = 50...subtract 45 from both sides

5/6x = 50 - 45

5/6x = 5.....divide both sides by 5/6 to get rid of the 5/6 on the left

x = 5 / (5/6)

x = 5 * 6/5

x = 30/5 which reduces to 6 <===

either way the answer is the same...x = 6...just do what ur most comfortable with

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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

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

Plug in x = 2 to find that...

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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,

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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.

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