Triangle, not a triangle, not a triangle, triangle
Determining if three side lengths can make a triangle is easier than it looks. All you have to do is use the Triangle Inequality Theorem, which states that the sum of two side lengths of a triangle is always greater than the third side. If this is true for all three combinations of added side lengths, then you will have a triangle.[
Answer:
She spent $33 dollars
Step-by-step explanation:
3 times 2.75 equals 8.25
4 times 5.50 equals 22
1 times 2.75 equals 2.75
8.25+22+2.75= $33
Answer:
Step-by-step explanation:
First, let us find the gradient of AB:
Gradient of AB = 
= 
We also need to know that The <em>product of gradients which are perpendicular to each other is -1</em>. Using this idea, we can find the gradient of the perpendicular bisector:
(Gradient of perpendicular bisector)( ) = -1
Gradient of perpendicular bisector = 
Now, we need to know at which coordinates the perpendicular bisector intersects AB. <em>A perpendicular bisector bisects a line to two equal parts</em>. Hence the <em>coordinates of the intersection point is the midpoint of AB</em>. Thus,
Coordinates of intersection = ( 
, 
)
= ( 2, 1 )
Now, we can construct our equation. The equation of a line can be formed using the formula 
where 
is the gradient and the line passes through 
. Hence by substituting the values, we get:
<u />
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>
D (1,4)
A (4,1)
C (0,6)
Hope this helps
