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Kazeer [188]
3 years ago
9

PLEASE HELP

Mathematics
1 answer:
devlian [24]3 years ago
7 0

Answer:

I think it is a 160o Angle

Step-by-step explanation:

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F plus 4 divided by h= 6
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Maurice and Johanna have appreciated the help you have provided them and their company Pythgo-grass. They have decided to let yo
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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
<span>
The squared distance from (x,y) to (p,q) is </span>(x-p)^2+(y-q)^2.<span>
These are equal in a parabola:

</span>
(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.

<span>3.There are two fruit trees located at (3,0) and (−3, 0) in the backyard plan. Maurice wants to use these two fruit trees as the focal points for an elliptical flowerbed. Johanna wants to use these two fruit trees as the focal points for some hyperbolic flowerbeds. Create the location of two vertices on the y-axis. Show your work creating the equations for both the horizontal ellipse and the horizontal hyperbola. Include the graph of both equations and the focal points on the same coordinate plane.

4.A pipe needs to run from a water main, tangent to a circular fish pond. On a coordinate plane, construct the circular fishpond, the point to represent the location of the water main connection, and all other pieces needed to construct the tangent pipe. Submit your graph. You may do this by hand, using a compass and straight edge, or by using a graphing software program.

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3 years ago
For which pairs of functions is (fOg)=12x
Komok [63]
G(x) = 3x 
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Airida [17]

Answer: ABC seem to be equilateral

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I=45

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Step-by-step explanation:

6 0
3 years ago
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Please someone help I don’t understand this
irina [24]

Answer:

336

Step-by-step explanation:

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3 years ago
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