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3 years ago

Domain and range of graphs

Mathematics

1 answer:

White raven [17]3 years ago

4 0

Answer:the domain is every “x” coordinate and the range is the “y” coordinate

Step-by-step explanation:

If you have to tell whether it’s a function or not, all of the x values in the domain will be different for a function, and 2 or more of the same x value won’t be a function

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Hello today is my birthday please help !!

scoray [572]

Answer:

the last one(y=2(2/3)^x) is the correct answer

Step-by-step explanation:

I identify two coordinate on the graph (0,2) and (1,3) and I noticed only the last one gives you a appropriate output if you plug the correspond input value

4 0

2 years ago

A scale map says that 1 cm represents 10 km. What distance on the map (in centimeters) represents an actual distance of 360 km?

castortr0y [4]

Answer:

36 cm

Step-by-step explanation:

1cm = 10 km

360km/10km=36

1x36=36

8 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

Help me plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz tired

LenKa [72]

So, y-intercept is (0,3) and x-intercept is (3/7,0)

Step-by-step explanation:

We are given:

$y=-7x+3$

We need to find y-intercept and x-intercept

Finding y-intercept

To find y-intercept we put x =0

Solving:

$y=-7x+3\\y=-7(0)+3\\y=3$

So, y-intercept is (0,3)

Finding x-intercept

To find x-intercept we put y =0

Solving:

$y=-7x+3\\0=-7x+3\\-7x=-3\\x=-3/-7\\x=3/7$

So, x-intercept is (3/7,0)

So, y-intercept is (0,3) and x-intercept is (3/7,0)

Keywords: x and y intercepts

Learn more about x and y intercepts at:

#learnwithBrainly

6 0

3 years ago

A widget company produces 25 widgets a day,5 of which are defective.Find the probability of selecting 5 widgets from the 25 prod

MariettaO [177]

Answer:

0.292

Step-by-step explanation:

Combinations can be used to solve the following problem.

We are choosing 5 widgets from 25 = ²⁵C₅

We want to select zero widgets from defective widgets = ⁵C₀

From the 20 non-defective widgets we want to select 5 = ²⁰C₅

So the probability is:

P = ( ⁵C₀ * ²⁰C₅) / ²⁰C₅

P = (1 * 15504) / 53130

= 15504/ 53130

=0.292 ..

7 0

3 years ago