(100-2x)\2 =,<,or> 4x(x + 1

Summarize what is a stock?

Naya [18.7K]

Answer:

well the definition shortened would be : to have or keep a supply of.

Step-by-step explanation:

3 0

3 years ago

What is the exact area of a circle with a diameter of 10 cm? a. 5πcm^2 b. 10πm^2 c. 25πcm^2 d. 100πcm^2

dalvyx [7]

Answer:

C

Step-by-step explanation:

a= $\pi$ r^2

To find the radius, divide the diameter by 2

r=d/2

r=10/2

r=5

a= $\pi$ r^2

a= $\pi$ 5^2

a=25 $\pi$

So, choice C is correct

Hope this helps! :)

4 0

3 years ago

56÷(7-9)^3 -24/23-5×4

yulyashka [42]

Answer:

= −28.043478261

Step By Step Explanation

Answer From Gauth Math

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

XYZ has coordinates X(2, 3), Y(1,4), and Z(8,9). A translation maps X to X'(4,7). What are the coordinates for Y' and Z' for thi

Rashid [163]

Answer:

The coordinates are $Y'(x,y) = (3, 8)$ and $Z'(x,y) = (10, 13)$ .

Step-by-step explanation:

First, we have to derive an expression for translation under the assumption that each point of XYZ experiments the same translation. Vectorially speaking, translation from X to X' is defined by:

$X'(x,y) = X(x,y) + T(x,y)$ (1)