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2 years ago

Help please!!!!!!!!!

Mathematics

1 answer:

finlep [7]2 years ago

4 0

Kill me please I'm ugly

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What is the radius of a circle with an area of 66 yards squared? Round to the nearest tenth of a yard.

Leto [7]

Answer: The answer is 0.41 or .41

Step-by-step explanation:

Move the

6 0

3 years ago

Read 2 more answers

Divide.<br> -6:31<br> Enter your answer as a mixed number, in simplified form, in the box.

jekas [21]

Answer:−

-12/31

Step-by-step explanation:

8 0

3 years ago

Seven friends went out to lunch. The meal subtotal was $91.22, including tax. The restaurant has a policy that for parties of si

vlada-n [284]

Answer:

$ 74.80

Step-by-step explanation:

The subtotal cost of the meal was $91.22 including tax.

The 18% tip is added for six or more.

The tip added should reduce the meal subtotal by 18 %

Reduce $91.22 by 18% to get ;

82/100 * 91.22 = $ 74.80

7 0

3 years ago

Im stomped with 42/12 in simple form. Im trying to remember if I should divide by 7.

Yuliya22 [10]

Exact form- 7/2
Decimal form- 3.5
Mixed number form- 3 1/2

I hope that helps!!!

5 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