Answer:
The intersection of the horizontal plane with the prism would have a square shape.
Step-by-step explanation:
A right rectangular prism is one in which each side makes an angle of 
with its base, which is a square. It is a three dimensional figure with six faces.
The point where the horizontal plane and the prism coincides would take the shape of the base of the prism. This is because the prism rests with its base on the horizontal plane. Thus, the shape required is in the form of a square since its base is a square. The length and width of its base are equal.
- The Midpoint of AB is (1,0).
Given that:
- In line AB, where the coordinates of A is (3,1) and coordinates of B is (-1,-1).
To find:
So, according the question
We know that,
The midpoint M of a line segment AB with endpoints A (x₁, y₁) and B (x₂, y₂) has the coordinates M ( 
).
Now from question,
We know that the the coordinates of A is (3,1) and coordinates of B is (-1,-1) of line AB.
So, we can say that
A is (3,1) or x₁ = 3 and y₁ = 1.
B (-1,-1) or x₂ = -1 and y₂ = -1.
∵ The coordinates of midpoint M (X,Y)
X = 
= 
= 2/2
X = 1.
And
Y = 
= 
= 0/2
Y = 0.
So, the midpoint of line AB is M (1,0)
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Answer:
k = y/x
Step-by-step explanation:
1/5, 2/10, 3/15 . . . all equal to 1/5 which is what the value of k is; 'k' stands for constant of proportionality
Answer:
Choice D is correct
Step-by-step explanation:
The eccentricity of the conic section is 1, implying we are looking at a parabola. Parabolas are the only conic sections with an eccentricity of 1.
Next, the directrix of this parabola is located at x = 4. This implies that the parabola opens towards the left and thus the denominator of its polar equation contains a positive cosine function.
Finally, the value of k in the numerator is simply the product of the eccentricity and the absolute value of the directrix;
k = 1*4 = 4
This polar equation is given by alternative D
Answer:
(3, - 5 )
Step-by-step explanation:
Under a clockwise rotation about the origin of 270°
a point (x, y ) → (- y, x ), hence
Z(5, - 3 ) → Z'(3, - 5 )
