Jenna bought 6 pounds of apples for $7.38. What is the unit price per pound of apples?

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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

When baking a cake, you have a choice of the following pans: a round cake pan that is 2 inches deep and has a 7 inch diameter a

ruslelena [56]

Round pan volume is:
3.14•r^2•h
D=7 so r=3.5 in
3.14• (3.5^2)•2 = 76.97 in^3

Rec. pan vol. is :
9•6•2= 108 in^3

Rec. Pan is larger because 108 in^3 is > 76.97 in^3 :) .

The icing that will be needed to frost the round cake pan is:
We need to find the surface area:
S.A= 3.14r^2 + 2 • 3.14•r • h .... 3.14 is the value of PI
So, S.A= 3.14• 3.5^2 + 6.28• 3.5• 2= 82.47 in^2 the icing that'll be needed to frost the round cake pan.

Icing that will be needed for the rec. cake pan is:

2•9•2=36 in^2
6•9•2= 108in^2
6•2= 12 in^2
S.A= 156 in^2 the icing needed to frost the rec. cake pan .... the S.A of all sides except the bottom one :).

Good luck ;-)

8 0

3 years ago

If a bakery produces 1,040 cupcakes during an 8 hour shift, what is the production rate per hour of cupcakes?

Fittoniya [83]

Answer: D

Step-by-step explanation: You do 1,040 divided by 8=130

Hope this helps!!!

8 0

4 years ago

Read 2 more answers

Which expressions are equal to 7(53)? Select all that apply.

skad [1K]

Answer:

7(50 + 3)

7(50) + 7(3)

371

Step-by-step explanation:

7(53)

50 can be written as (50 + 3)

∴7(53) = 7(50 + 3)

Proceeding with the above expression we get;

7(53) = 7(50 + 3)

= 7(50) + 7(3)

Completing the above expression we get;

7(53) = 7(50) + 7(3)

= 350 + 21

= 371

3 0

3 years ago

Daphne has 3 days to drive 932.4 miles . she wants to have completed about 75% of the drive by the start of day three.which of t

nekit [7.7K]

So, 932.4miles (.75)=699.3miles are the miles that she wants to drive by the start of three days. so each day is 25% because of 75%/3=25%
In two days it will be 50% so, .50(932.4miles)=466.2 miles is the answer.

4 0

3 years ago