Cindy and Zoey work at a store. Both girls earn $6.25 per hour

Mathematics

1 answer:

wolverine [178]3 years ago

8 0

Answer:

Step-by-step explanation:

I think this is your full question, right?

Cindy and Zoey work at a store. Both girls earn $6.25 per hour. During a normal week Cindy works 15 hours and Zoey works 20 hours. The expression 6.25(15) + 6.25(20) can be used to calculate the total amount of money that the girls earned in one week. Which expression shows another way to calculate the amount of money the girls earn in one week?

My answer:

The expression shows another way must be equal to: 6.25(15) + 6.25(20)

<=> 6.25 (15+20)

Because both girls earn the same per hour.

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A gallon jug of milk is 5/6 full. After breakfast the jug is 1/12 full. Find the difference of the amounts before breakfast and

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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

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

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3 years ago

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nataly862011 [7]

Answer:

the characters choices often cause the conflicts

create links to other slides , videos or documents

a presentation that does not have any interaction

Step-by-step explanation:

hope I helped!! ;)

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