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Alex17521 [72]
4 years ago
11

Does someone understand this ?

Mathematics
1 answer:
Nimfa-mama [501]4 years ago
3 0
<h3>ANSWERS:</h3>

1. Line XY

2. Line XZ

3. (opposite angle): IU

(adjacent angle): IH

Question 1:

For the first problem, you are finding the opposite side to ∠YZX, by knowing that Z is the center of this angle, it is rather simple to solve.

Just draw your line through the middle of the angle through the triangle until you hit the side on the other side.

When doing this on question 1, you would pass between the angles of ∠X and ∠Y, <u><em>therefore you are passing through line XY.</em></u>

Question 2:

For the next problem, we need the adjacent angle to ∠YZX, in other words the one other line which is not the opposite angle.

By doing that we can rule it down to lines XZ and YZ, but it cannot be angle YZ because that is the hypotenuse of the triangle (the longest side) and cannot ever be the opposite or adjacent lines.

<u><em>So, your answer for question 2 would be line XZ.</em></u>

Question 3:

For the last problem, we need to find both the opposite and adjacent sides of the triangle according to angle ∠IHU.

So, firstly, do that finger through the center of the angle trick for the opposite angle and you would get line IU.

Then for the adjacent angle, find the hypotenuse and choose the one other angle that you haven't chosen as your opposite angle.

By doing that, you would get line IH.

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The average daily temperature, t, in degrees fahrenheit for a city as a function of the month of the year, m, can be modeled by
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The answer will be t = sin((m.\frac{\pi}{6})+55) average daily temperature, t, in degrees Fahrenheit for a city as a function of the month of the year.

<h3>What is temperature?</h3>

Temperature is the degree or intensity of heat present in a substance or object, especially as expressed according to a comparative scale and shown by a thermometer or perceived by touch.

<h3>TO SOLVE:</h3>

35cos(\frac{\pi}{6(m+3)}+55)\\\\35 cos(\frac{\pi m}{6} + 55 + \frac{\pi}{2})\\

suppose x = m.\frac{\pi}{6}+55 and \frac{\pi}{2} = 90 degree

We know, cos(x+90°) = - sin(X)

⇒ cos((m.\frac{\pi}{6})+55+90degree)degree = -sin((m.\frac{\pi}{6})+55)

⇒ t = sin((m.\frac{\pi}{6})+55)

Hence the answer will be t = sin((m.\frac{\pi}{6})+55) average daily temperature, t, in degrees Fahrenheit for a city as a function of the month of the year.

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