Q1. The answer is x = 1, y = 1, z = 0
<span>(i) -2x+2y+3z=0
</span><span>(ii) -2x-y+z=-3
</span>(iii) <span>2x+3y+3z=5
</span><span>_________
Sum up the first and the third equation:
</span>(i) -2x+2y+3z=0
(iii) 2x+3y+3z=5
_________
5y + 6z = 5
Sum up the second and the third equation:
(ii) -2x-y+z=-3
(iii) 2x+3y+3z=5
_________
2y + 4z = 2
(iv) 5y + 6z = 5
(v) 2y + 4z = 2
________
Divide the fifth equation by 2
(iv) 5y + 6z = 5
(v) y + 2z = 1
________
Multiple the second equation by -3 and sum the equation
(iv) 5y + 6z = 5
(v) -3y - 6z = -3
________
2y = 2
y = 2/2 = 1
y + 2z = 1
1 + 2z = 1
2z = 1 - 1
2z = 0
z = 0
-2x-y+z=-3
-2x - 1 + 0 = -3
-2x = -3 + 1
-2x = -2
x = -2/-2 = 1
Q2. The answer is x = -37, y = -84, z = -35
<span>(i) x-y-z=-8
(ii) -4x+4y+5z=7
(iii) 2x+2z=4
______
</span>Divide the third equation by 2 and rewrite z in the term of x:
(iii) x+z=2
z = 2 - x
______
Substitute z from the third equation and express y in the term of x:
<span>x-y-(2-x)=-8
x - y - 2 + x = 8
2x - y = 10
y = 2x - 10
______
Substitute z from the third equation and y from the first equation into the second equation:
</span><span>-4x + 4y + 5z = 7
-4x + 4(2x - 10) + 5(2 - x) = 7
-4x + 8x - 40 + 10 - 5x = 7
-x -30 = 7
-x = 30 + 7
x = -37
y = 2x - 10 = 2*(-37) - 10 = -74 - 10 = -84
z = 2 - x = 2 - 37 = -35</span>
Answer:
or 2/3
Step-by-step explanation:
because two flips out of 3 all together is 2/3
Missing information:
How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point
![P = (\frac{1}{2}, \frac{\sqrt 3}{2})](https://tex.z-dn.net/?f=P%20%3D%20%28%5Cfrac%7B1%7D%7B2%7D%2C%20%5Cfrac%7B%5Csqrt%203%7D%7B2%7D%29)
Answer:
![Rate = 0.935042^\circ /cm](https://tex.z-dn.net/?f=Rate%20%3D%200.935042%5E%5Ccirc%20%2Fcm)
Step-by-step explanation:
Given
![P = (\frac{1}{2}, \frac{\sqrt 3}{2})](https://tex.z-dn.net/?f=P%20%3D%20%28%5Cfrac%7B1%7D%7B2%7D%2C%20%5Cfrac%7B%5Csqrt%203%7D%7B2%7D%29)
![T(x,y) =x\sin2y](https://tex.z-dn.net/?f=T%28x%2Cy%29%20%3Dx%5Csin2y)
![r = 1m](https://tex.z-dn.net/?f=r%20%3D%201m)
![v = 2m/s](https://tex.z-dn.net/?f=v%20%3D%202m%2Fs)
Express the given point P as a unit tangent vector:
![u = \frac{\sqrt 3}{2}i - \frac{1}{2}j](https://tex.z-dn.net/?f=u%20%3D%20%5Cfrac%7B%5Csqrt%203%7D%7B2%7Di%20-%20%5Cfrac%7B1%7D%7B2%7Dj)
Next, find the gradient of P and T using: ![\triangle T = \nabla T * u](https://tex.z-dn.net/?f=%5Ctriangle%20T%20%3D%20%5Cnabla%20T%20%2A%20u)
Where
![\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})} = (sin \sqrt 3)i + (cos \sqrt 3)j](https://tex.z-dn.net/?f=%5Cnabla%20T%7C_%7B%28%5Cfrac%7B1%7D%7B2%7D%2C%20%5Cfrac%7B%5Csqrt%203%7D%7B2%7D%29%7D%20%20%3D%20%28sin%20%5Csqrt%203%29i%20%2B%20%28cos%20%5Csqrt%203%29j)
So: the gradient becomes:
![\triangle T = \nabla T * u](https://tex.z-dn.net/?f=%5Ctriangle%20T%20%3D%20%5Cnabla%20T%20%2A%20u)
![\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j]](https://tex.z-dn.net/?f=%5Ctriangle%20T%20%3D%20%5B%28sin%20%5Csqrt%203%29i%20%2B%20%28cos%20%5Csqrt%203%29j%5D%20%2A%20%20%5B%5Cfrac%7B%5Csqrt%203%7D%7B2%7Di%20-%20%5Cfrac%7B1%7D%7B2%7Dj%5D)
By vector multiplication, we have:
![\triangle T = (sin \sqrt 3)* \frac{\sqrt 3}{2} - (cos \sqrt 3) \frac{1}{2}](https://tex.z-dn.net/?f=%5Ctriangle%20T%20%3D%20%28sin%20%5Csqrt%203%29%2A%20%20%5Cfrac%7B%5Csqrt%203%7D%7B2%7D%20-%20%28cos%20%5Csqrt%203%29%20%20%5Cfrac%7B1%7D%7B2%7D)
![\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)](https://tex.z-dn.net/?f=%5Ctriangle%20T%20%3D%200.9870%20%2A%200.8660%20-%20%28-0.1606%20%2A%200.5%29)
![\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5](https://tex.z-dn.net/?f=%5Ctriangle%20T%20%3D%200.9870%20%2A%200.8660%20%2B0.1606%20%2A%200.5)
![\triangle T = 0.935042](https://tex.z-dn.net/?f=%5Ctriangle%20T%20%3D%200.935042)
Hence, the rate is:
Answer:
I don't know if you're asking for the product of prime using index notation but I got :
2 × 2 × 2 × 5 × 5
Answer:
oh my- this actually made me happy-
Step-by-step explanation: