Answer:
12
Step-by-step explanation:
43
Answer:
9 bags
Step-by-step explanation:
First we will need to calculate the area of their entire backyard, which has a shape of trapezium.
So the formula for the area of a trapezium is 
a and b are the lenghts of the parallel sides and h is the height.
From the picture 
Then we plug into our formula
square feet
To get the number of bags required we just simply divide the aree of their entire yard by the space that 1 bag covers. We get:
bags
This is more than 8 bags so they will need 9 bags to cover their entire yard.
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
Answer:
t ≤ 15
Step-by-step explanation:
One third of 15 is equal to five (1/3 × 15 = 5)
One third of t is lesser than or equal to five (1/3 × 5 ≤ 5)
<em>So, t cannot be more than 15.</em><em> </em><em>It</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>equal</em><em> </em><em>to</em><em> </em><em>1</em><em>5</em><em>,</em><em> </em><em>though</em><em>.</em>
∴t ≤ 15