Answer:
If you take 32 percent of a number and get 64, then what is that number? In other words, you know that 32 percent of a number is 64 and you want to know what that initial number is. Therefore, the answer to "64 is 32 percent of what number?" is 200, and you can also derive that 32 percent of 200 equals 64.
Step-by-step explanation:
The correct answer choice is A. I think!
Try this option:
1. note, that the number 84 is divided by numbers: 1;2;3;4;6;7;12;14;21;28;42 and 84. It means, that measurements of the length may be: 1;2;3;4;6;7;12;14;21;28;42;84 and for the width: 84;42;28;21;14;12;7;6;4;3;2;1.
2. according the numbers calculated in item 1, the possible number measurements for the length and width of the sandbox are (length; width):
1;84
2;42
3;28
4;21
6;14
7;12
12;7
14;6
21;4
28;3
42;2
84;1
Answer:
Step-by-step explanation:
<u>Given function:</u>
The function is plotted and it is below.
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>
