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andreev551 [17]
3 years ago
15

Is Figure A congruent to Figure B? Explain.

Mathematics
1 answer:
velikii [3]3 years ago
6 0
Yes, because figure B can be obtained from figure A by a series of translations
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Second to last one is my guess
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<span><span>1.A triangular section of a lawn will be converted to river rock instead of grass. Maurice insists that the only way to find a missing side length is to use the Law of Cosines. Johanna exclaims that only the Law of Sines will be useful. Describe a scenario where Maurice is correct, a scenario where Johanna is correct, and a scenario where both laws are able to be used. Use complete sentences and example measurements when necessary.
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The Law of Cosines is always preferable when there's a choice.  There will be two triangle angles (between 0 and 180 degrees) that share the same sine (supplementary angles) but the value of the cosine uniquely determines a triangle angle.

To find a missing side, we use the Law of Cosines when we know two sides and their included angle.   We use the Law of Sines when we know another side and all the triangle angles.  (We only need to know two of three to know all three, because they add to 180.  There are only two degrees of freedom, to answer a different question I just did.

<span>2.An archway will be constructed over a walkway. A piece of wood will need to be curved to match a parabola. Explain to Maurice how to find the equation of the parabola given the focal point and the directrix.
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We'll use the standard parabola, oriented in the usual way.  In that case the directrix is a line y=k and the focus is a point (p,q).

The points (x,y) on the parabola are equidistant from the line to the point.  Since the distances are equal so are the squared distances.

The squared distance from (x,y) to the line y=k is </span>(y-k)^2
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The squared distance from (x,y) to (p,q) is </span>(x-p)^2+(y-q)^2.<span>
These are equal in a parabola:

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(y-k)^2 =(x-p)^2+(y-q)^2<span>

</span>y^2-2ky + k^2 =(x-p)^2+y^2-2qy + q^2

y^2-2ky + k^2 =(x-p)^2 + y^2 - 2qy+ q^2

2(q-k)y =(x-p)^2+ q^2-k^2

y = \dfrac{1}{2(q-k)} ( (x-p)^2+ q^2-k^2)

Gotta go; more later if I can.

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How do you graph the equation Y = 4x + 2 and 4x - 3y = 12?
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Answer:look below

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Answer:

A

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