All categories

3 years ago

The set of order pairs (-1, 8), (0,3) (1,2) and (2, -7) represent a function. What is the range of the function?

1 answer:

8 0

Answer: The range is the set {8, 3, 2, -7} which can be sorted to get {-7, 2, 3, 8}

------------------------------------------------------------------------------

Explanation:

The range is simply the set of possible y values. All you have to do is list all the y coordinates and surround the values in curly braces. The curly braces indicate a set. Every point is of the form (x,y). The y coordinate is the second value listed of each ordered pair.

You might be interested in

Suppose you work for a life insurance company. At your job, you use the following table, which shows the number of people out of

irina [24]

Oh boi I'm gonna fail, I can't even geuss

3 0

3 years ago

Read 2 more answers

Answer this question for please!Describe the steps.Thanks!

GREYUIT [131]

1. 75/300 = x/48 (your proportion)
2. x= 12

3. 30/240 = x/24 (your proportion)
4. x= 3

Your answer would come out to 15 miles total.

4 0

3 years ago

Plz helpppppppppppppppppppppppppppppppppppppppppppppppppppppp

BigorU [14]

D is the answer . . .

8 0

3 years ago

Read 2 more answers

Ab=45 mm,bc=10cm 4mm ,cd = 5 cm 6 mm ,de= 35 mm,ea=x cm, and the perimeter of irregular pentagon is 280 mm. Find the length of t

DaniilM [7]

Answer:

The length of the unknown side x is 4 cm

Step-by-step explanation:

we know that

The perimeter of a polygon is equal to the sum of its length sides

Remember that

1 cm=10 mm

In this problem we have

AB=45 mm ------> convert to cm -----> AB=45/10=4.5 cm

BC=10 cm 4 mm -----> convert to cm -----> BC=10+4/10=10.4 cm

CD=5 cm 6 mm -----> convert to cm -----> CD=5+6/10=5.6 cm

DE=35 mm ------> convert to cm -----> DE=35/10=3.5 cm

EA=x cm

P=280 mm ------> convert to cm -----> P=280/10=28 cm

The perimeter of the irregular polygon is equal to

P=AB+BC+CD+DE+EA

substitute the values

28=4.5+10.4+5.6+3.5+x

28=24+x

x=28-24

x=4 cm

3 0

3 years ago

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

3 years ago