Answer:
I got A.
Step-by-step explanation:
Answer:
256.
Step-by-step explanation:
Each position can be one of 2 states so the number of possibilities
is 2^8 = 256.
Answer:
Two boxes will cost
The cost of one tile is approximately ![\dfrac{\$71.9}{67}\approx \$1.08.](https://tex.z-dn.net/?f=%5Cdfrac%7B%5C%2471.9%7D%7B67%7D%5Capprox%20%5C%241.08.)
Step-by-step explanation:
Note that 1 ft = 12 in, then
![7\ ft\ 6\ in=7\cdot 12+6\ in =90\ in,\\ \\8\ ft\ 10\ in=8\cdot 12+10\ in=106\ in.](https://tex.z-dn.net/?f=7%5C%20ft%5C%206%5C%20in%3D7%5Ccdot%2012%2B6%5C%20in%20%3D90%5C%20in%2C%5C%5C%20%5C%5C8%5C%20ft%5C%2010%5C%20in%3D8%5Ccdot%2012%2B10%5C%20in%3D106%5C%20in.)
The area of the floor is
![90\cdot 106=9540\ in^2.](https://tex.z-dn.net/?f=90%5Ccdot%20106%3D9540%5C%20in%5E2.)
Now
![1\ ft^2=144\ in^2,](https://tex.z-dn.net/?f=1%5C%20ft%5E2%3D144%5C%20in%5E2%2C)
then
144 sq. in - 1 tile,
9540 sq. in - x tiles.
Mathematically,
![\dfrac{144}{9540}=\dfrac{1}{x},\\ \\x=\dfrac{9540}{144}=66.25](https://tex.z-dn.net/?f=%5Cdfrac%7B144%7D%7B9540%7D%3D%5Cdfrac%7B1%7D%7Bx%7D%2C%5C%5C%20%5C%5Cx%3D%5Cdfrac%7B9540%7D%7B144%7D%3D66.25)
tiles is needed to cover the floor (67 full tiles).
Each box of tiles contains 45 tiles, so Cristina has to buy 2 boxes. These two boxes will cost
The cost of one tile is approximately ![\dfrac{\$71.9}{67}\approx \$1.08.](https://tex.z-dn.net/?f=%5Cdfrac%7B%5C%2471.9%7D%7B67%7D%5Capprox%20%5C%241.08.)
For the answer to the question above,
<span>r = 1 + cos θ
x = r cos θ
x = ( 1 + cos θ) cos θ
x = cos θ + cos^2 θ
dx/dθ = -sin θ + 2 cos θ (-sin θ)
dx/dθ = -sin θ - 2 cos θ sin θ
y = r sin θ
y = (1 + cos θ) sin θ
y = sin θ + cos θ sin θ
dy/dθ = cos θ - sin^2 θ + cos^2 θ
dy/dx = (dy/dθ) / (dx/dθ)
dy/dx = (cos θ - sin^2 θ + cos^2 θ)/ (-sin θ - 2 cos θ sin θ)
For horizontal tangent line, dy/dθ = 0
cos θ - sin^2 θ + cos^2 θ = 0
cos θ - (1-cos^2 θ) + cos^2 θ = 0
cos θ -1 + 2 cos^2 θ = 0
2 cos^2 θ + cos θ -1 = 0
Let y = cos θ
2y^2+y-1=0
2y^2+2y-y-1=0
2y(y+1)-1(y+1)=0
(y+1)(2y-1)=0
y=-1
y=1/2
cos θ =-1
θ = π
cos θ =1/2
θ = π/3 , 5π/3
θ = π/3 , π, 5π/3
when θ = π/3, r = 3/2
when θ = π, r = 0
when θ = 5π/3 , r = 3/2
(3/2, π/3) and (3/2, 5π/3) give horizontal tangent lines
</span>---------------------------------------------------------------------------------
For horizontal tangent line, dx/dθ = 0
<span>-sin θ - 2 cos θ sin θ = 0 </span>
<span>-sin θ (1+ 2 cos θ ) = 0 </span>
<span>sin θ = 0 </span>
<span>θ = 0, π </span>
<span>(1+ 2 cos θ ) =0 </span>
<span>cos θ =-1/2 </span>
<span>θ = 2π/3 </span>
<span>θ = 4π/3 </span>
<span>θ = 0, 2π/3 ,π, 4π/3 </span>
<span>when θ = 0, r=2 </span>
<span>when θ = 2π/3, r=1/2 </span>
<span>when θ = π, r=0 </span>
<span>when θ = 4π/3 , r=1/2 </span>
<span>(2,0) , (1/2, 2π/3) , (0, π), (1/2, 4π/3) </span>
<span>At (2,0) there is a vertical tangent line</span>