Is Y=-3/4x -2 Y=3/4x+1 One solution many solution or no solution

bulgar [2K]

-2,3 is (x1,y1) and (4,-5) is (x2,y2) so you do y2-y1/x2-x1

y2-y1 is -5-3 which is -8 and x2-x1 is 4-(-2) which is 6 so you get

-8/6= -4/3

4 0

2 years ago

Derivative, by first principle <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

the ages of the 11 players of a soccer team on the soccer field are shown. 21, 32, 22, 28, 26, 30, 29, 24, 22, 24, 26. what woul

horrorfan [7]

If a smaller number was added as an outlier

7 0

2 years ago

The sum of the first 20 positive odd integers

svet-max [94.6K]

Answer:

$400$

Step-by-step explanation:

$We\ know\ that,\\'The\ sum\ of\ n\ no.\ positive\ odd\ integers\ is\ given\ by\ the\ formula:\\S_O=n^2\\Hence,\\Here,\\As\ n=20,\\The\ sum\ of\ the\ first\ 20\ positive\ odd\ integers=20^2=400\\$

7 0

2 years ago

A wildlife reserve had 8 elephant calves born during the summer and now has 31 total elephants. How many elephants were in the r

Elza [17]

Because you now have 31 you - 8 from 31
equation 31-8=e
solve 23 elephant before summer

6 0

2 years ago