Answer:
5/2cups
Step-by-step explanation:
If a recipe serves 6 and calls for 3/4cups, arecipe that serves 20, will call for (20×3/4)/6 cups. Mathematically;
6recipe = 3/4cups
20recipe = x
Cross multiplying we have;
6x = 20×3/4
6x = 15
x = 15/6
x = 5/2
This means a recipe that serves 20 will call for 5/2cups
Answer:
![\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos%282x%29%7D%20-%20%5Csqrt%5B3%5D%7Bcos%283x%29%7D%7D%7Bsin%28x%5E2%29%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D)
General Formulas and Concepts:
<u>Calculus</u>
Limits
Limit Rule [Variable Direct Substitution]: 
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)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28g%28x%29%29%5D%20%3Df%27%28g%28x%29%29%20%5Ccdot%20g%27%28x%29)
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)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos%282x%29%7D%20-%20%5Csqrt%5B3%5D%7Bcos%283x%29%7D%7D%7Bsin%28x%5E2%29%7D)
When we directly plug in <em>x</em> = 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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos%282x%29%7D%20-%20%5Csqrt%5B3%5D%7Bcos%283x%29%7D%7D%7Bsin%28x%5E2%29%7D%20%3D%20%5Cfrac%7B0%7D%7B0%7D)
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)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos%282x%29%7D%20-%20%5Csqrt%5B3%5D%7Bcos%283x%29%7D%7D%7Bsin%28x%5E2%29%7D%20%3D%20%5Cdisplaystyle%20%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5Cfrac%7B-sin%282x%29%7D%7B%5Csqrt%7Bcos%282x%29%7D%7D%20%2B%20%5Cfrac%7Bsin%283x%29%7D%7B%5Bcos%283x%29%5D%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%7D%7B2xcos%28x%5E2%29%7D)
Plugging in <em>x</em> = 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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5Cfrac%7B-sin%282x%29%7D%7B%5Csqrt%7Bcos%282x%29%7D%7D%20%2B%20%5Cfrac%7Bsin%283x%29%7D%7B%5Bcos%283x%29%5D%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%7D%7B2xcos%28x%5E2%29%7D%20%3D%20%5Cfrac%7B0%7D%7B0%7D)
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)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5Cfrac%7B-sin%282x%29%7D%7B%5Csqrt%7Bcos%282x%29%7D%7D%20%2B%20%5Cfrac%7Bsin%283x%29%7D%7B%5Bcos%283x%29%5D%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%7D%7B2xcos%28x%5E2%29%7D%20%3D%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5Cfrac%7B-%5Bcos%5E2%282x%29%20%2B%201%5D%7D%7B%5Bcos%282x%29%5D%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%20%2B%20%5Cfrac%7Bcos%5E2%283x%29%20%2B%202%7D%7B%5Bcos%283x%29%5D%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%7D%7D%7B2cos%28x%5E2%29%20-%204x%5E2sin%28x%5E2%29%7D)
Substitute in <em>x</em> = 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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5Cfrac%7B-%5Bcos%5E2%282x%29%20%2B%201%5D%7D%7B%5Bcos%282x%29%5D%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%20%2B%20%5Cfrac%7Bcos%5E2%283x%29%20%2B%202%7D%7B%5Bcos%283x%29%5D%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%7D%7D%7B2cos%28x%5E2%29%20-%204x%5E2sin%28x%5E2%29%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D)
And we have our final answer.
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Limits
6a perimeter is 12 cms and 28cms.
6b side length is 90 cms
6c areas 9sq cm and 49 sq cms
6d 11 cms
Answer:
9 in
Step-by-step explanation:
For an n-sided polygon, the length of the apothem is ...
a = r·cos(180°/n)
We assume your problem statement is saying the radius is 10 inches. For a hexagon, n=6 and we have ...
a = (10 in)cos(30°) ≈ 8.66 in
Rounded to the nearest inch, the apothem is 9 in.
(4,-2) (6,-2) (4,1) are co ordinates for the triangle after the movement