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3 years ago

Match each expression to the scenario it represents.

Mathematics

2 answers:

pickupchik [31]3 years ago

3 0

Answer:

A = A

F = B

I don't know of the others

Step-by-step explanation:

hammer [34]3 years ago

3 0

The person above is correct with those 2 sadly I can’t help either with the others

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I need halp PLSSSSS I’m confused

vfiekz [6]

Answer:

the answer is C (18:30)

Step-by-step explanation:

Each set of ratio in the table in simplest form is 3:5, and so is C.

6×3=18

6×5=30

So, 18:30 in simplest form is 3:5

Hope this helps!

4 0

3 years ago

Jorge earns $9,567 a month. Which number is $9,567 rounded to the nearest hundred?

Maksim231197 [3]

To find $9,567 rounded to the nearest hundred you must first find the hundreds place which is where 5 is then according to the number behind it you either round up or you keep 5 the same since the number behind 5 is 6 you round 5 up one which brings 5 to 6 now everything behind your new number is turned to 0. so your new amount would be $9,600

7 0

3 years ago

A castle has to be guarded 24 hours a day. Five knights are ordered to split each day’s guard duty equally. How long will each k

Kisachek [45]

I think that is 248 min in one day because 24/5 is 4.8. 4hours and 8 min. 4 times 60 is 240. 240 plus 8 is 248 this is what I think.

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

3 years ago

sharia has a full time job and earns 475$ each week but she starts looking for ways to increase her weekly earnings . sharia fin

Arlecino [84]

Answer:

12 hours

Step-by-step explanation:

Full time job:

Earning per week = $475

Part time job :

Earning per hour = $11

Number of working hours per week at per time job to increase weekly earning to atleast $600 ;

Let number of hours = h

Full time earning + earning per hour * (number of hours) ≥ 600

$475 + ($11 * h) ≥ 600

$475 + $11h ≥ 600

11h ≥ 600 - 475

11h ≥ 125

h ≥ 125 / 11

h ≥ 11.36

Hence, sharia will need to work for 12 hours per week

5 0

3 years ago