Answer: <span>(x, y) → (x + 8, y – 3)
Justification:
</span>1) From the figure you can tell that the original trapezoid ABCD was translated 8 units to the right and 3 units down to yield the trapezoid A'B'C'D'.
2) Translating 8 units to the right means: x → x + 8
3) Translating 3 units down means y → y - 3
4) Therefore, the pair (x,y) is transformed into the pair (x + 8, y - 3); that is the rule (x,y) → (x + 8, y - 3)
Answer:
would go for C, SSS (side side side theorem)
He would need 2.8 bouquet's to make 20 dollars because i had to round
i hope this helped
Answer:
Step-by-step explanation:
From the given information:
r = 10 cos( θ)
r = 5
We are to find the the area of the region that lies inside the first curve and outside the second curve.
The first thing we need to do is to determine the intersection of the points in these two curves.
To do that :
let equate the two parameters together
So;
10 cos( θ) = 5
cos( θ) = ![\dfrac{1}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D)
![\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20-%5Cdfrac%7B%5Cpi%7D%7B3%7D%2C%20%5C%20%5C%20%20%5Cdfrac%7B%5Cpi%7D%7B3%7D)
Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e
![A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta](https://tex.z-dn.net/?f=A%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cint%20%5Climits%5E%7B%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D_%7B-%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D%20%2810%20%5C%20cos%20%5C%20%20%5Ctheta%29%5E2%20d%20%5Ctheta%20-%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cint%20%5Climits%5E%7B%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D_%7B-%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D%20%5C%20%5C%20%205%5E2%20d%20%5Ctheta)
![A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta](https://tex.z-dn.net/?f=A%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cint%20%5Climits%5E%7B%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D_%7B-%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D%20100%20%5C%20cos%5E2%20%5C%20%20%5Ctheta%20%20d%20%5Ctheta%20-%20%5Cdfrac%7B25%7D%7B2%7D%20%5Cint%20%5Climits%5E%7B%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D_%7B-%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D%20%5C%20%5C%20%20%20d%20%5Ctheta)
![A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}](https://tex.z-dn.net/?f=A%20%3D%2050%20%5Cint%20%5Climits%5E%7B%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D_%7B-%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D%20%5Cbegin%20%7Bpmatrix%7D%20%20%5Cdfrac%7Bcos%20%5C%202%20%5Ctheta%20%2B1%7D%7B2%7D%20%20%5Cend%20%7Bpmatrix%7D%20%5C%20%5C%20d%20%5Ctheta%20-%20%5Cdfrac%7B25%7D%7B2%7D%20%20%5Cbegin%20%7Bbmatrix%7D%20%5Ctheta%20%20%20%5Cend%20%7Bbmatrix%7D%5E%7B%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D_%7B-%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D)
![A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}](https://tex.z-dn.net/?f=A%20%3D%5Cdfrac%7B%2050%7D%7B2%7D%20%5Cint%20%5Climits%5E%7B%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D_%7B-%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D%20%5Cbegin%20%7Bpmatrix%7D%20%20%7Bcos%20%5C%202%20%5Ctheta%20%2B1%7D%20%20%5Cend%20%7Bpmatrix%7D%20%5C%20%5C%20%20%20%20d%20%5Ctheta%20-%20%5Cdfrac%7B25%7D%7B2%7D%20%20%5Cbegin%20%7Bbmatrix%7D%20%20%5Cdfrac%7B%5Cpi%7D%7B3%7D%20-%20%28-%20%5Cdfrac%7B%5Cpi%7D%7B3%7D%20%29%5Cend%20%7Bbmatrix%7D)
![A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}](https://tex.z-dn.net/?f=A%20%3D25%20%20%5Cbegin%20%7Bbmatrix%7D%20%20%5Cdfrac%7Bsin2%20%5Ctheta%20%7D%7B2%7D%20%2B%20%5Ctheta%20%5Cend%20%7Bbmatrix%7D%5E%7B%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D_%7B%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D%20%20%20%20%5C%20%5C%20-%20%5Cdfrac%7B25%7D%7B2%7D%20%20%5Cbegin%20%7Bbmatrix%7D%20%20%5Cdfrac%7B2%20%5Cpi%7D%7B3%7D%20%5Cend%20%7Bbmatrix%7D)
![A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}](https://tex.z-dn.net/?f=A%20%3D25%20%20%5Cbegin%20%7Bbmatrix%7D%20%20%5Cdfrac%7Bsin%20%28%5Cdfrac%7B2%20%5Cpi%7D%7B3%7D%20%29%7D%7B2%7D%2B%5Cdfrac%7B%5Cpi%7D%7B3%7D%20-%20%5Cdfrac%7B%20sin%20%28%5Cdfrac%7B-2%5Cpi%7D%7B3%7D%29%20%7D%7B2%7D-%28-%5Cdfrac%7B%5Cpi%7D%7B3%7D%29%20%20%5Cend%20%7Bbmatrix%7D%20-%20%5Cdfrac%7B25%20%5Cpi%7D%7B3%7D)
![A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}](https://tex.z-dn.net/?f=A%20%3D%2025%20%5Cbegin%7Bbmatrix%7D%20%20%20%5Cdfrac%7B%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%20%7D%7B2%7D%20%2B%5Cdfrac%7B%5Cpi%7D%7B3%7D%20%2B%20%5Cdfrac%7B%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%20%7D%7B2%7D%20%2B%20%20%20%5Cdfrac%7B%5Cpi%7D%7B3%7D%20%20%5Cend%20%7Bbmatrix%7D-%20%5Cdfrac%7B%2025%20%5Cpi%7D%7B3%7D)
![A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}](https://tex.z-dn.net/?f=A%20%3D%2025%20%5Cbegin%7Bbmatrix%7D%20%20%20%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%20%7D%20%2B%5Cdfrac%7B2%20%5Cpi%7D%7B3%7D%20%20%20%5Cend%20%7Bbmatrix%7D-%20%5Cdfrac%7B%2025%20%5Cpi%7D%7B3%7D)
![A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}](https://tex.z-dn.net/?f=A%20%3D%20%20%20%20%5Cdfrac%7B25%20%5Csqrt%7B3%7D%7D%7B2%20%7D%20%2B%5Cdfrac%7B25%20%5Cpi%7D%7B3%7D)
The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.
<em>If the number you are rounding is followed by 5, 6, 7, 8, or 9, round the number up (+1).</em>
<em>If the number you are rounding is followed by 0, 1, 2, 3, or 4, round the number down (no change).</em>
81.1<u>3</u>9 ≈ <u>81.1</u>