Answer:
Sinla
Step-by-step explanation:
Kanya 0.15
Than three-twentieths = 3/20 = .15
Sinla two-ninths = 2/9 = .22222
Jessica 0.2
Divide your fractions to get a decimal. Compare your decimals.
The surface (call it
) is a triangle with vertices at the points
![x=0,y=0\implies z=2\implies(0,0,2)](https://tex.z-dn.net/?f=x%3D0%2Cy%3D0%5Cimplies%20z%3D2%5Cimplies%280%2C0%2C2%29)
![x=0,y=2\implies z=-2\implies(0,2,-2)](https://tex.z-dn.net/?f=x%3D0%2Cy%3D2%5Cimplies%20z%3D-2%5Cimplies%280%2C2%2C-2%29)
![x=2,y=0\implies z=-8\implies(2,0,-8)](https://tex.z-dn.net/?f=x%3D2%2Cy%3D0%5Cimplies%20z%3D-8%5Cimplies%282%2C0%2C-8%29)
Parameterize
by
![\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)](https://tex.z-dn.net/?f=%5Cvec%20s%28u%2Cv%29%3D%281-v%29%282%2C0%2C-8%29%2Bv%5Cbigg%28%281-u%29%280%2C2%2C-2%29%2Bu%280%2C0%2C2%29%5Cbigg%29%3D%282-2v%2C2v-2uv%2C-8%2B6v%2B4uv%29)
with
and
. Take the normal vector to
to be
![\vec s_v\times\vec s_u=(20v,8v,4v)](https://tex.z-dn.net/?f=%5Cvec%20s_v%5Ctimes%5Cvec%20s_u%3D%2820v%2C8v%2C4v%29)
Then the flux of
across
is
![\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_S%5Cvec%20F%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20d%5Cvec%20S%3D%5Cint_0%5E1%5Cint_0%5E1%5Cvec%20F%28x%28u%2Cv%29%2Cy%28u%2Cv%29%2Cz%28u%2Cv%29%29%5Ccdot%28%5Cvec%20s_v%5Ctimes%5Cvec%20s_u%29%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv)
![=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E1%5Cint_0%5E1%286-6v%2C2v-2uv%2C-8%2B6v%2B4uv%29%5Ccdot%2820v%2C8v%2C4v%29%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv)
![=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle8%5Cint_0%5E1%5Cint_0%5E1%2811v-10v%5E2%29%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv%3D%5Cboxed%7B%5Cfrac%7B52%7D3%7D)
Average=(total number)/(number of items)
given that the final exam counts as two test, let the final exam be x. The weight of the final exams on the average is 2, thus the final exam can be written as 2x because any score Shureka gets will be doubled before the averaging.
Hence our inequality will be as follows:
(67+68+76+63+2x)/6≥71
(274+2x)/6≥71
solving the above we get:
274+2x≥71×6
274+2x≥426
2x≥426-274
2x≥152
x≥76
b] The above answer is x≥76, the mean of this is that if Shureka is aiming at getting an average of 71 or above, then she should be able to get a minimum score of 76 or above. Anything less than 76 will drop her average lower than 71.
Its the 3, 4, 5 Pythagorean Triple so it would be D 8 inches