Answer:
Option B. 1 : 3
Step-by-step explanation:
There are two triangles shown in the picture attached ΔABC and ΔCD.
In these triangles sides AB and DE are parallel and BC is transverse line so ∠ABC = ∠EDC (corresponding angles)
Similarly AC is transverse to parallel lines AB and DE, so ∠BAC = ∠DEC (corresponding angles)
∠ACB is common in both the triangles.
Therefore ΔABC and ΔDEC are similar.
We know in similar triangles corresponding sides are in the same ratio.
Therefore ratio of y and 12 is equal to 3 : 1
Or ratio of 12 to y is 1 : 3
Option B is the answer.
Answer:
hi kenma-san!!!!!!
Step-by-step explanation:
A trapezoid with height 6.4 cm and parallel base is 12.9 cm and 8.6 cm.
We have to find the area of area of this trapezoid.
Area of trapezoid =
Given : Height = 6.4 cm
and parallel base is 12.9 cm and 8.6 cm.
Substitute, we have,
Area of trapezoid =
Simplify, we have,
Area of trapezoid =
Area of trapezoid = 68.8 cm²
Thus, The area of given trapezoid is 68.8 cm²
2.2 maybe? I would check with someone
Answer:
yes
Step-by-step explanation:
The line intersects each parabola in one point, so is tangent to both.
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For the first parabola, the point of intersection is ...
y^2 = 4(-y-1)
y^2 +4y +4 = 0
(y+2)^2 = 0
y = -2 . . . . . . . . one solution only
x = -(-2)-1 = 1
The point of intersection is (1, -2).
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For the second parabola, the equation is the same, but with x and y interchanged:
x^2 = 4(-x-1)
(x +2)^2 = 0
x = -2, y = 1 . . . . . one point of intersection only
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If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.
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Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.
I think its times 10 *shrug*
