Rotating a figure 90° counterclockwise about the origin results in an ordered pair (x, y) mapping to its image (-y, x). Thus point C, which is at (3, 5), will map to (-5, 3).
Answer: Work done by the particle is 78 N.
Step-by-step explanation:
If C is the path the particle follows, then work done is 
.
According to the question the force 
and all four points are in the plane 
.
therefore, if S is the at surface with boundary C, so that S is the portion of the plane over the rectangle D=[0,2]\times [0,4].
Now,
By the Stoke's theorem:
![\int_{C}^{}F.dx=\iint_{s}^{}curlF.dS\\\\=\iint_{D}^{}[-8y(0)-2z\left ( \frac{1}{4} \right )+5y]dA\\\\=\iint_{D}^{}\left ( -\frac{1}{2}z+5y \right )dA](https://tex.z-dn.net/?f=%5Cint_%7BC%7D%5E%7B%7DF.dx%3D%5Ciint_%7Bs%7D%5E%7B%7DcurlF.dS%5C%5C%5C%5C%3D%5Ciint_%7BD%7D%5E%7B%7D%5B-8y%280%29-2z%5Cleft%20%28%20%5Cfrac%7B1%7D%7B4%7D%20%5Cright%20%29%2B5y%5DdA%5C%5C%5C%5C%3D%5Ciint_%7BD%7D%5E%7B%7D%5Cleft%20%28%20-%5Cfrac%7B1%7D%7B2%7Dz%2B5y%20%5Cright%20%29dA)
![\int_{0}^{2}\left [ \frac{39}{16}y^2 \right ]_{0}^{4}dx\\\\=\int_{0}^{2}39\, \, dx\\\\=39\times 2\\\\=78 \, N](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7B2%7D%5Cleft%20%5B%20%5Cfrac%7B39%7D%7B16%7Dy%5E2%20%5Cright%20%5D_%7B0%7D%5E%7B4%7Ddx%5C%5C%5C%5C%3D%5Cint_%7B0%7D%5E%7B2%7D39%5C%2C%20%5C%2C%20%20dx%5C%5C%5C%5C%3D39%5Ctimes%202%5C%5C%5C%5C%3D78%20%5C%2C%20N)
Answer:
1. 3-2 6-5 9-8 18-17 21-20
2. sharri
3. no because 5 x 3 = 15 and 7 x 3 does not equal 17
4. 22.5
5. 110 miles
hope this helps :)
5n² - 8n - 21 → (5)(- 21) = -105 (find 2 numbers whose product is 105 and sum is -8)
= 5n² - 15n + 7n - 21
= 5n(n - 3) + 7(n - 3)
= (5n + 7)(n - 3)
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6n² + 53n - 79 → (6)( -79) = -474 (find 2 numbers whose product is -474 and sum is 53) DOES NOT EXIST! only factors are: (1, 474) (2, 237) (3, 158) (6, 79)
= Prime
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4n² - 25 (this is the difference of two squares)
= √(4n²) +/- √(25)
= 2n +/- 5
= (2n + 5)(2n - 5)
Answer:
x=4± 5sqrt(2)
Step-by-step explanation:
x^2 – 8x – 34 = 0
To complete the square Add 34 to each side
x^2 -8x -34+34=0+ 34
Take the coefficient of x, and divide by 2
-8/2 =-4
Then square it and add it to each side
(-4)^2 =16
x^2 – 8x +16 = 34+16
x^2 – 8x +16 = 50
We replace the left side with (x + the coefficient of x/2)^2
(x -4)^2=50
Take the square root of each side
sqrt((x -4)^2)=±sqrt(50)
x-4 = ±sqrt(25*2)
x-4 = ±sqrt(25)*sqrt(2)
x-4 = ±5sqrt(2)
Add 4 to each side
x=4± 5sqrt(2)
