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3 years ago

Which of the following ordered pairs could be placed in the table and still have the relation qualify as a linear function?

Mathematics

2 answers:

Svetradugi [14.3K]3 years ago

8 0

Answer:

Step-by-step explanation:

hammer [34]3 years ago

7 0

The answer is B, by the way don't listen to the guy above me, because 1,7 is completely wrong. the answer is (2, 13), because the table is showing that for every value of x the output, (y), changes by 3, and according to the table,we need to add three more which equals thirteen, then we move the x value up.

I hope this helps!

btw feel free to subscribe to my channel vexus gamer with a sans icon on it, because memes.

You might be interested in

-0.125 as a fraction

kaheart [24]

Answer:

-1/8

Step-by-step explanation.

125 x 8 = 1000

.125 x 8 = 1

-.125 x 8 = -1

-1/8

8 0

3 years ago

Answer these 2 questions for 100 pts. answers only please

aliya0001 [1]

Answer:

%K = 31.904%

%Cl = 28.930%

%O = 39.166%

Hg = 80.69/ 200.59 = 0.40/0.2 = 2

S = 6.436/ 32.07 = 0.20/0.2 = 1

O = 12.87/ 16.00 = 0.80/0.2 = 4

Hg2SO4 is the empirical formula

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

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

4 years ago

Help help help plz plz plz

DaniilM [7]

Answer:

Yes

Step-by-step explanation:

What would you do if he said ok so he said yes would go?

I told him god bless him and to keep working in his vocabulary.

5 0

3 years ago

The side of a cube is measured to be 5.1cm. Between what limits of reading does its volume lie? State the limits to 3 decimal pl

Aliun [14]

132.651 cm bc you do 5.1 x 5.1 x5. 1 and you will get 132.651 cm.
volume=l x w x h

6 0

3 years ago