All categories

2 years ago

Please help me with number 12!

Mathematics

1 answer:

Olegator [25]2 years ago

4 0

The answer is 10 quarters because 10 quarters would be $2.50. You would need 3 more nickles to get $2.65. 10 would be 7 more than 3.

You might be interested in

<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

2 years ago

Akiba Ellis deposits the following in her checking account: 6 five-dollar bills, 5 two dollar bills, 12 one dollar bills, 9 half

In-s [12.5K]

Answer:

$106.08

Step-by-step explanation:

Given:

6x $5.00

5x $2.00

12x $1.00

9x= 0.50

6x25 cent

42x10 cent

10x5 cent

15=15cent

$97.23

----------------------------------------------------------------------------------

6 five dollar bill is $30.00

so 5 two dollar bill is $10.00

12 one dollar bills is $12.00

9 half dollar bill mean 0.50 cent since it Half and that would make $4.50

6 quarter so 6x25 which equal $1.50

42 dimes and dimes =10 cent so 42x10 which equal $4.20

10 nickels and nickels = 5 cent so 10x5=0.50

15=15 pennies

-------------------------------------------------------------------------------------

30.00+10.00+12.00+4.50+1.50+4.20+0.50+0.15+97.23 =

$106.08 Total Deposit

5 0

2 years ago

Dalia bought a few swirl marbles and divided it equally among four of her friends

Gennadij [26K]

Answer:

16 ang answer

Step-by-step explanation:

Kung mali comment me

6 0

1 year ago

What is 5.75% of $6,000? (Express answer as dollars as cents.)

Len [333]

6000*0.0575=345
Final answer: $345.00

6 0

2 years ago

Read 2 more answers

Help with 2 algebra questions? Will give lots of points too.

Nat2105 [25]

1 b
2 c
Hope it helps:)

3 0

2 years ago

Read 2 more answers