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xeze [42]
3 years ago
11

If a point has a negative x-coordinate, and its y-coordinate is positive, where does the point lie relative to the origin? Expla

in how you know.
Mathematics
1 answer:
Daniel [21]3 years ago
4 0
      If a coordinate has a negative x-coordinate and its y-coordinate is positive, then the point  is on the 2nd quadrant.
      That means that the point is left of the origin and above it.
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34 minus 20 equals 14

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One angle of a triangle measures 136°. The other two angles are congruent. Enter and solve an equation to find the measure x of
Juliette [100K]

<u>136 + x + x = 180. Or, to simplify, 136 + 2x = 180. The congruent angles measure 23 degrees each.</u>

We know this because of a simple rule that goes for all triangles: The measures of all three angles in a triangle will <em>always</em> add up to 180 degrees.

One angle of a triangle measures 136 degrees. The other two angles are congruent (have the same measure).

(x stands for an unknown angle measure.) So the equation we would use is 136 + 2x = 180. We can solve this within a few steps.

1. We subtract 136 from 2x in order to isolate 2x. But if we subtract something from the left side of the equation, we have to subtract it from the right side too. Otherwise the equation will be wrong; We would be taking away the balance.

2x = 180 - 136

2. Now that 2x is isolated, we solve 180 - 136.

2x = 46

3. If we know now that 2x is equal to 46, how do we find out what x is equal to? We divide by 2 (on both sides or it'll be wrong) to get x.

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x = 23

Now we know! x = 23... The other two angles are both 23 degrees. We can check to see if that's right by solving 23 + 23 + 136... Does it add up to 180? Yes! :)

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3 years ago
Cos(x+y)/cosxcosy+tanxtany=1
adelina 88 [10]

Answer:

See explanation

Step-by-step explanation:

Use formula

\cos (x+y)=\cos x\cos y-\sin x\sin y

Substitute it into the first fraction:

\dfrac{\cos (x+y)}{\cos x\cos y}\\ \\=\dfrac{\cos x\cos y-\sin x\sin y}{\cos x\cos y}\\ \\=\dfrac{\cos x\cos y}{\cos x\cos y}-\dfrac{\sin x\sin y}{\cos x\cos y}\\ \\=1-\dfrac{\sin x}{\cos x}\cdot \dfrac{\sin y}{\cos y}\\ \\=1-\tan x\tany

Consider the whole expression:

\dfrac{\cos (x+y)}{\cos x\cos y}+\tan x\tan y\\ \\=1-\tan x \tan y+\tan x\tan y\\ \\=1

Done!

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