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3 years ago

What is the range of magazines sold? please answer will give brainliest

Mathematics

2 answers:

timurjin [86]3 years ago

3 0

Answer:

Step-by-step explanation:

wariber [46]3 years ago

3 0

Answer:

range: 8

Step-by-step explanation:

In order to find the range you need the minimum and maximum

minimum: 17

maximum: 25

Subtract

25-17=8

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ANYONE PLEASE HELP ME IN MY MATH HOMEWORK I REALLY NEED THE ANSWER RIGHT NOW BECAUSE I HAVE TO PASS THIS TOMORROW:(

vladimir1956 [14]

Answer:

.....................

10)=(-3)^2.x^4×2.y^2(5.x^2y^6)-7^2x^5×2y^4×2

=(9×5)x^8+2y^2+6-49x^10y^8

=45x^8+2y^2+6-49x^10y^8

=45x^10y^8-49x^10y^8

=-4x^10y^8

5 0

3 years ago

hafizah harvested 49 mangoes from her farm. The weights of the mangoes, w in grams, are shown on the following frequency table

sergejj [24]

To get the answer for this problem your going to have to show us the frequency table

5 0

3 years 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.