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3 years ago

What two numbers multiply to 630 and add to -57

Mathematics

2 answers:

Mila [183]3 years ago

6 0

Well, see, you are looking for two negative numbers. So: -15 and -42.

sveta [45]3 years ago

5 0

-15 and -42 both add to -57 and multiply to 630

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2 years ago

Sue Jones is insured for bodily injury in the amount of 10/20. She is at fault in an accident in which Albert Smith is injured.

Vlad1618 [11]

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3 years ago

How do three planes intersect at one point?

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3 years ago

singh works at a restaurant where he is paid by the hour plus time and a half for his hours over 40. last week he worked four ho

padilas [110]

Answer:

$8.5 per hour

Step-by-step explanation:

Number of overtime hours = 4 hours

Since overtime starts after 40 hours, then the number of hours worked = 40 + 4 = 44 hours

Total amount earned = $391

Let Hourly pay = x

Overtime pay = 1 1/2x = 3/2x

Therefore ;

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40x + 6x = 391

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x = 391 / 46

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7 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