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Alecsey [184]
3 years ago
14

How do you find three consecutive even integers with a sum of -84. Please help!

Mathematics
2 answers:
mote1985 [20]3 years ago
6 0
3 consecutive even integers : x, x + 2, x + 4

x + (x + 2) + (x + 4) = -84.....combine like terms
3x + 6 = - 84.......subtract 6 from each side
3x = -84 - 6
3x = - 90...divide both sides by 3
x = -90/3
x = -30

x + 2 = -30 + 2 = -28
x + 4 = -30 + 4 = -26

so ur 3 numbers are : -26, -28, -30
matrenka [14]3 years ago
4 0
Since they are consecutive even integers, they would differ by 2.

Let the first even integer be x.

x , x + 2,  x + 4

Sum = -84

x + x + 2 + x + 4 = -84

x + x + x + 2 + 4 = -84

3x + 6 = -84

3x = -84 - 6

3x = -90

x = -90/3

x = -30

Hence the even integers are:  x,  x + 2 , x + 4

-30,  -30 + 2,  -30 + 2 + 2

-30,  -28, -26

Hence the consecutive even integers are: -30,  -28, -26
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Answer:

I would say C not to sure tho

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1 year 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.

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