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Naddika [18.5K]
3 years ago
10

Which equation best represents the graph?

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
3 0

Answer:

The third one

Step-by-step explanation:

The graphs in the reversed shape of a cup are the graphs of quadratic equations whose coefficient of

x^2 is negative.

So, the answer must be either the first or the third.

It can be seen that when x= -4, y becomes 5.

Substituting the value of x, the third equation gives the value 5.

So, the answer is the third one.

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Pls help asap for 15 points easy 3(2a–1)−5a =
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Answer:

a-3

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3(2a–1)−5a =

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20,25,30,30,31,40,41,49

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Anton [14]

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Arc ECG is the major arc

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3 years ago
Maurice and Johanna have appreciated the help you have provided them and their company Pythgo-grass. They have decided to let yo
Colt1911 [192]
<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.
</span>
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.
</span>
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>
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</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

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

Gotta go; more later if I can.

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