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2 years ago

I dance I sing I am perfect I am I am a believer that I am a discoverer. PLEASE ANSWER I'LL MARK AND BRAINLIEST

Mathematics

2 answers:

Viefleur [7K]2 years ago

7 0

Answer:

YAZZ You are amazing :> remind yourself everyday!!!

Julli [10]2 years ago

5 0

Answer:

not sure if its a song or not but you do you, gg <3

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2 4/5 divided by 2/3

daser333 [38]

Answer:

4.2 cuz yeah

Yeet :)

5 0

3 years ago

Read 2 more answers

2. Lamar purchases comic books at garage

Otrada [13]

Answer:

Lamar will charge $\$7$ to the comic dealer for a comic book.

Solution:

It is given that lamar purchases a comic book for $\$4$ . He then marks up the price of a comic book by $75\%$ which means he increases the price of the book by $75\%$ and then sells it to the comic dealer.

We have to find out how much will he charge the comic book dealer.

New price of the book after lamar marks up the price is given by

New price $= \$4 + \frac{75}{100} \times \$4$

= $\$4 + \$3$ $= \$7$

Thus, lamar will charge $\$7$ to the comic dealer for a comic book.

4 0

3 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

I need help quick please

dsp73

96 ! If you get x on one side and the # on the other you get x=96

4 0

3 years ago

The pink paper for the book Giver

Leni [432]

I think its the promise to not share it with anyone idk

4 0

3 years ago