Step-by-step explanation:
Given the ratio 5:3
and Points A and B where A is located at (-6, 3), and B is located at (26, -13).
Let point A be 
, and let point B be 
.
(-6, 3) → .
(26, -13) → .
Let 5 be n, and 3 be m.
5:3 → n:m
.
To solve, just substitute these variables into the expressions of these coordinates to get the answer.
→
→
→
→
→
→
→
.
Thus the coordinates of B are:
D=100:
g=12-1/25(100)
g=12-4
g=8, d=100
d=200:
g =12-1/25(200)
g=12-8
g=4, d=200
d=300:
g =12-1/25(300)
g=12- 12
g=0, d=300
Answer: True
Step-by-step explanation:
Substitute -1 for x and 1 for y in the equation y = 4x + 5
(1) = 4(-1) + 5
1 = -4 + 5
1 = 1
Left side = right side
Therefore, the point (–1, 1) satisfies the equation y = 4x + 5
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>
C = 14π cm ( in terms of π )
the circumference (C) of a circle is calculated using the formula
C = πd ( where d is the diameter )
C = 14π cm ≈ 3.14 × 14 - 43.96 cm
