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krek1111 [17]
3 years ago
6

Can someone solve 5 + 2x = + 6 and n - 3n = 14 - 4n and 7 ( 5a - 4 ) - 1 = 14 - 8a

Mathematics
2 answers:
Vladimir79 [104]3 years ago
8 0

Step-by-step explanation:

1.

5 + 2x = + 6

2x=6-5

2x=1

x \ =  \frac{1}{2}

2.

n - 3n = 14 - 4n

-2n= 14 -4n

-2n+4n=14

2n=14

n=7

3.

7(5a - 4 ) - 1 = 14 - 8a

35a-28-1=14-8a

35a-29=14-8a

35a+8a=14+29

43a=43

a= 1

professor190 [17]3 years ago
6 0

Answer:

5 + 2x = 6

2x = 6-5

2x = 1

x = 0.5

n -3n = 14 -4n

-2n = 14-4n (since 4n is negative, change the sign to make it positive and then add 4n to both sides of the equation)

2n = 14

n = 7

7 (5a - 4) - 1 = 14 - 8a

35a - 28 - 1 = 14 -8a

35a -29 = 14 - 8a

43a = 43

a = 1

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Calculus piecewise function. ​
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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

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
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