By Green's theorem,
![\displaystyle\int_C\cos y\,\mathrm dx+x^2\sin y\,\mathrm dy=\iint_D\left(\frac{\partial(x^2\sin y)}{\partial x}-\frac{\partial(\cos y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_C%5Ccos%20y%5C%2C%5Cmathrm%20dx%2Bx%5E2%5Csin%20y%5C%2C%5Cmathrm%20dy%3D%5Ciint_D%5Cleft%28%5Cfrac%7B%5Cpartial%28x%5E2%5Csin%20y%29%7D%7B%5Cpartial%20x%7D-%5Cfrac%7B%5Cpartial%28%5Ccos%20y%29%7D%7B%5Cpartial%20y%7D%5Cright%29%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy)
where
is the region with boundary
, so we have
![\displaystyle\iint_D(2x+1)\sin y\,\mathrm dx\,\mathrm dy=\int_0^5\int_0^4(2x+1)\sin y\,\mathrm dy\,\mathrm dx=\boxed{60\sin^22}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_D%282x%2B1%29%5Csin%20y%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%3D%5Cint_0%5E5%5Cint_0%5E4%282x%2B1%29%5Csin%20y%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dx%3D%5Cboxed%7B60%5Csin%5E22%7D)
Angle is half the arc
46/2= 23 C
Answer:
40cm² there you go
Step-by-step explanation:
Area =side×side so 8×5=40
<span>Let's call these friends by their first initial:
K = Kiana, G = Gary, M = Monty, W = Willa
Now we can convert the question to a system of linear equations:
Gary has 3 cards more than Kiana, so G = K + 3
Monty has 7 cards more than Kiana, so M = K + 7
Willa has twice as many cards as Monty, so W = 2M = 2(K + 7) = 2K + 14
74 cards given out = all the friends combined, so
74
= K + G + M + W
= K + (K + 3) + (K + 7) + (2K + 14)
= 5K + 3 + 7 + 14
= 5K + 24
50 = 5K
10 = K, so Kiana had 10 cards
G = K + 3 = 13, so Gary had 13 cards
M = K + 7 = 17, so Monty had 17 cards
W = 2M = 34, so Willa had 34 cards
Checking our work, 10 + 13 + 17 + 34 = 74 cards total.</span>
Answer:
x = (270° - 9y°)/2
y = (270° - 2x°)/9
Step-by-step explanation:
The sum of angles in a quadrilateral is 360°
The angles in the quadrilateral above are: 3y°, 6y°, 2x° and 90°
2x° + 3y° + 6y° + 90° = 360°
2x° + 9y° = 360° - 90°
2x° + 9y° = 270°
2x° = 270° - 9y°
x = (270° - 9y°)/2
2x° + 9y° = 270°
9y° = 270° - 2x°
y = (270 - 2x°)/9