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NARA [144]
4 years ago
7

Can I please get some help in numbers 2, 4, 8, 16, 38, and 40

Mathematics
1 answer:
alexgriva [62]4 years ago
6 0

Step-by-step explanation:

2. cross products

4. solve

8. 3 is to 8 as 10 is to 27

The proportion would be: 3/8 = 10/27

The cross products are 3 * 27 = 81 and 8 * 10 = 80

Since the cross products are not equal, it is not a proportion.

False.

16. The cross products are 5 * 2 = 10 and 30 * 1/3 = 10.

Since the cross products are equal, it is a proportion.

True.

38.

3/4 * x = 3 * 6

4/3 * 3/4 * x = 4/3 * 18

x = 24

40.

24 * x = 1.8 * 28

24x = 50.4

24x/24 = 50.4/24

x = 2.1

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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
2 years ago
For a school fundraiser students are selling snack bags and candy bars to raise money on Wednesday the students or 23 snack bags
meriva

Answer:

$1.75

Step-by-step explanation:

The selling for each candy bar may be determined by  a set of linear equations. This pair of linear equations may be solved simultaneously by using the elimination method. This will involve ensuring that the coefficient of one of the unknown variables is the same in both equations.

It may be solved by substitution in that one of the variable is made the subject of the equation and the result is substituted into the second equation .

Let the cost of a snack bag be s and that of a candy bar be c, then if on Wednesday the students or 23 snack bags and 36 candy bars that raised $114.75 on Thursday the seventh so 37 snack bags and 36 candy bars that raised $146.25

23s + 36c = 114.75

37s + 36c = 146.25

14s = 31.5

s = $2.25

23(2.25) + 36c = 114.75

36c = 114.75 - 51.75

36c = 63

c = 63/36

= $1.75

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3 years ago
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