Hello!
Answer:
To solve, we will need to use Right-Triangle Trigonometry:
Begin by solving for angles ∠S and ∠R using tangent (tan = opp/adj)
tan ∠S = a / (1/2b)
tan ∠S = 3√5 / 14
tan ∠S ≈ 0.479
arctan 0.479 = m∠S (inverse)
m∠S and m∠R ≈ 25.6°
Use cosine to solve for the hypotenuse, or the missing side-length:
cos ∠S = 14 / x
x · cos (25.6) = 14
x = 14 / cos(25.6)
x ≈ 15.52
Both triangles are congruent, so we can go ahead and find the perimeter of the figure:
RS + RQ + QS = 28 + 15.52 + 15.52 = 59.04 units.
Hope this helped you! :)
Question in English:
Tom has 2,021 gold coins. He divides them into piles containing consecutive numbers of pieces.
if it has more than two piles, how much is in the highest pile?
Plzzzzz give me Brainliest!!!!!
Answer:
First option: Square
with diagonals 
and 
.
Fifth option: Segment 
is parallel to segment 
.
Step-by-step explanation:
We know that a square is quadriteral whose sides are equal.
By definition, a square has the following properties:
1) Its four sides are congruent.
2) The diagonals are congruent.
3) The angles formed by the intersection of the diagonals measure 90 degrees.
4) The opposite sides are parallel.
5) Each internal angle measures 90 degrees.
Notice in the figure that:
- The diagonals of the square are: and .
- and are opposite sides, therefore, they are parallel.
Answer: 9
Step-by-step explanation: 105/12=8.75
8 batches are not enough so she should do it one more time to finish
Answer:
![\textsf{Midpoint rule}: \quad \dfrac{2\pi}{\sqrt[3]{2}}](https://tex.z-dn.net/?f=%5Ctextsf%7BMidpoint%20rule%7D%3A%20%5Cquad%20%5Cdfrac%7B2%5Cpi%7D%7B%5Csqrt%5B3%5D%7B2%7D%7D)
Step-by-step explanation:
<u>Midpoint rule</u>
![\displaystyle \int_{a}^{b} f(x) \:\:\text{d}x \approx h\left[f(x_{\frac{1}{2}})+f(x_{\frac{3}{2}})+...+f(x_{n-\frac{3}{2}})+f(x_{n-\frac{1}{2}})\right]\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7Ba%7D%5E%7Bb%7D%20f%28x%29%20%5C%3A%5C%3A%5Ctext%7Bd%7Dx%20%5Capprox%20h%5Cleft%5Bf%28x_%7B%5Cfrac%7B1%7D%7B2%7D%7D%29%2Bf%28x_%7B%5Cfrac%7B3%7D%7B2%7D%7D%29%2B...%2Bf%28x_%7Bn-%5Cfrac%7B3%7D%7B2%7D%7D%29%2Bf%28x_%7Bn-%5Cfrac%7B1%7D%7B2%7D%7D%29%5Cright%5D%5C%5C%5C%5C%20%5Cquad%20%5Ctextsf%7Bwhere%20%7Dh%3D%5Cdfrac%7Bb-a%7D%7Bn%7D)
<u>Trapezium rule</u>
![\displaystyle \int_{a}^{b} y\: \:\text{d}x \approx \dfrac{1}{2}h\left[(y_0+y_n)+2(y_1+y_2+...+y_{n-1})\right] \quad \textsf{where }h=\dfrac{b-a}{n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7Ba%7D%5E%7Bb%7D%20y%5C%3A%20%5C%3A%5Ctext%7Bd%7Dx%20%5Capprox%20%5Cdfrac%7B1%7D%7B2%7Dh%5Cleft%5B%28y_0%2By_n%29%2B2%28y_1%2By_2%2B...%2By_%7Bn-1%7D%29%5Cright%5D%20%5Cquad%20%5Ctextsf%7Bwhere%20%7Dh%3D%5Cdfrac%7Bb-a%7D%7Bn%7D)
<u>Simpson's rule</u>
<u>Given definite integral</u>:
![\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%5E%7B2%20%5Cpi%7D_0%20%5Csqrt%5B3%5D%7B%5Csin%5E2%20%28x%29%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx)
Therefore:
Calculate the subdivisions:
<u>Midpoint rule</u>
Sub-intervals are:
![\left[0, \dfrac{1}{2}\pi \right], \left[\dfrac{1}{2}\pi, \pi \right], \left[\pi , \dfrac{3}{2}\pi \right], \left[\dfrac{3}{2}\pi, 2 \pi \right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%20%5Cdfrac%7B1%7D%7B2%7D%5Cpi%20%5Cright%5D%2C%20%5Cleft%5B%5Cdfrac%7B1%7D%7B2%7D%5Cpi%2C%20%5Cpi%20%5Cright%5D%2C%20%5Cleft%5B%5Cpi%20%2C%20%5Cdfrac%7B3%7D%7B2%7D%5Cpi%20%5Cright%5D%2C%20%5Cleft%5B%5Cdfrac%7B3%7D%7B2%7D%5Cpi%2C%202%20%5Cpi%20%5Cright%5D)
The midpoints of these sub-intervals are:
Therefore:
![\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2}\pi \left[f \left(\dfrac{1}{4} \pi \right)+f \left(\dfrac{3}{4} \pi \right)+f \left(\dfrac{5}{4} \pi \right)+f \left(\dfrac{7}{4} \pi \right)\right]\\\\& = \dfrac{1}{2}\pi \left[\sqrt[3]{\dfrac{1}{2}} +\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}\right]\\\\ & = \dfrac{2\pi}{\sqrt[3]{2}}\\\\& = 4.986967483...\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cdisplaystyle%20%5Cint%5E%7B2%20%5Cpi%7D_0%20%5Csqrt%5B3%5D%7B%5Csin%5E2%20%28x%29%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%20%26%20%5Capprox%20%5Cdfrac%7B1%7D%7B2%7D%5Cpi%20%5Cleft%5Bf%20%5Cleft%28%5Cdfrac%7B1%7D%7B4%7D%20%5Cpi%20%5Cright%29%2Bf%20%5Cleft%28%5Cdfrac%7B3%7D%7B4%7D%20%5Cpi%20%5Cright%29%2Bf%20%5Cleft%28%5Cdfrac%7B5%7D%7B4%7D%20%5Cpi%20%5Cright%29%2Bf%20%5Cleft%28%5Cdfrac%7B7%7D%7B4%7D%20%5Cpi%20%5Cright%29%5Cright%5D%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Cpi%20%5Cleft%5B%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B2%7D%7D%20%2B%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B2%7D%7D%2B%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B2%7D%7D%2B%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B2%7D%7D%5Cright%5D%5C%5C%5C%5C%20%26%20%3D%20%5Cdfrac%7B2%5Cpi%7D%7B%5Csqrt%5B3%5D%7B2%7D%7D%5C%5C%5C%5C%26%20%3D%204.986967483...%5Cend%7Baligned%7D)
<u>Trapezium rule</u>
![\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2} \cdot \dfrac{1}{2} \pi \left[(0+0)+2(1+0+1)\right]\\\\& = \dfrac{1}{4} \pi \left[4\right]\\\\& = \pi\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cdisplaystyle%20%5Cint%5E%7B2%20%5Cpi%7D_0%20%5Csqrt%5B3%5D%7B%5Csin%5E2%20%28x%29%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%20%26%20%20%5Capprox%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ccdot%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cpi%20%5Cleft%5B%280%2B0%29%2B2%281%2B0%2B1%29%5Cright%5D%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7B1%7D%7B4%7D%20%5Cpi%20%5Cleft%5B4%5Cright%5D%5C%5C%5C%5C%26%20%3D%20%5Cpi%5Cend%7Baligned%7D)
<u>Simpson's rule</u>
<u />
<u />![\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(0+4(1)+2(0)+4(1)+0\right)\\\\& = \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(8\right)\\\\& = \dfrac{4}{3} \pi\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cdisplaystyle%20%5Cint%5E%7B2%20%5Cpi%7D_0%20%5Csqrt%5B3%5D%7B%5Csin%5E2%20%28x%29%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%20%26%20%5Capprox%20%5Cdfrac%7B1%7D%7B3%7D%5Ccdot%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cpi%20%5Cleft%280%2B4%281%29%2B2%280%29%2B4%281%29%2B0%5Cright%29%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7B1%7D%7B3%7D%5Ccdot%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cpi%20%5Cleft%288%5Cright%29%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7B4%7D%7B3%7D%20%5Cpi%5Cend%7Baligned%7D)
