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lidiya [134]
3 years ago
11

Tell how you can critique Jason reasoning

Mathematics
2 answers:
NeTakaya3 years ago
4 0
Is there another part to your question
Natasha2012 [34]3 years ago
4 0
Is that all you got (Savage)
You might be interested in
What is 30=12m I need help
Law Incorporation [45]
30=12m
you divide both sides by 12
so 30/12 is 2.5
8 0
3 years ago
Calculando a expressão (-2)² + (+5)², obtemos qual resultado?
Contact [7]
This answer would be D) + 6.


Hope this helps if so please mark me as brainlest thank you !!
7 0
2 years ago
Read 2 more answers
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
1 year ago
Is the correctx 25&lt;28
trapecia [35]

Answer:

Yes

Step-by-step explanation:

25 is less than 28

28 is greater than 25

4 0
3 years ago
Read 2 more answers
In the diagram below,AB is the diameter of the circle with centre P.Point C lies on the y-axis.Given the coordinates of A and B
Step2247 [10]

Answer:

<em>The coordinates of P are (-2,0)</em>

<em>The radius of the circle is 5.</em>

Step-by-step explanation:

<u>Analytic geometry</u>

The diagram shows a circle with center P, and two points A(-6,3) and B(2,-3) that form the diameter of the circle.

a)

The center of the circle lies at the midpoint of A and B. The midpoint (xm,ym) can be calculated by:

\displaystyle x_m=\frac{x_1+x_2}{2}

\displaystyle y_m=\frac{y_1+y_2}{2}

Substituting x1=-6, x2=2, y1=3, y2=-3:

\displaystyle x_m=\frac{-6+2}{2}=\frac{-4}{2}=-2

\displaystyle y_m=\frac{3-3}{2}=0

Thus, the coordinates of P are (-2,0)

b) The radius of the circle is the distance from the center to any point in its circumference. We can use the distance from P to A or B indistinctly.

Given two points A(x1,y1) and P(x2,y2), the distance between them is:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Substituting x1=-6, x2=-2, y1=3, y2=0:

r=\sqrt{(-2+6)^2+(0-3)^2}

r=\sqrt{4^2+(-3)^2}

r=\sqrt{16+9}=\sqrt{25}=5

The radius of the circle is 5.

4 0
3 years ago
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