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LenaWriter [7]
3 years ago
15

Which step is included in the construction of parallel lines?

Mathematics
1 answer:
tekilochka [14]3 years ago
6 0
<span>Copy an angle by creating arcs with a compass.

</span>
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Given csc(A) = 60/16 and that angle A is in Quadrant I, find the exact value of sec A in simplest radical form using a rational
Katarina [22]

Answer:

\displaystyle \sec A=\frac{65}{63}

Step-by-step explanation:

We are given that:

\displaystyle \csc A=\frac{65}{16}

Where <em>A</em> is in QI.

And we want to find sec(A).

Recall that cosecant is the ratio of the hypotenuse to the opposite side. So, find the adjacent side using the Pythagorean Theorem:

a=\sqrt{65^2-16^2}=\sqrt{3969}=63

So, with respect to <em>A</em>, our adjacent side is 63, our opposite side is 16, and our hypotenuse is 65.

Since <em>A</em> is in QI, all of our trigonometric ratios will be positive.

Secant is the ratio of the hypotenuse to the adjacent. Hence:

\displaystyle \sec A=\frac{65}{63}

6 0
3 years ago
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Help me please, brainliested ;)
jenyasd209 [6]

Answer:

put a dot on -5 on the y axis

Step-by-step explanation:

3 0
3 years ago
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Please help me with these
BigorU [14]

go yfihuygfffccijj sghhuusfhhjdugkuyyggtnntguutujj

3 0
3 years ago
Nevermind anyone know this answer???
Semenov [28]

Answer:

x = 6

Step-by-step explanation:

for SPECIAL RIGHT TRIANGLES specifically a 30°-60°-90° triangle.

The longest side is the hypotenuse. The longest side is twice the shortest side. Or the other way around, the shortest side is half the longest side.

Here the longest side (hypotenuse) is 12, so the shortest side is 6.

Also, since you seem to be trying to get this subject.

The long leg is the short leg× sqroot3.

OR the short leg is the long leg ÷ sqroot3. If you end up with a sqroot on the bottom of a fraction, you can't leave it like that. Times top and bottom by sqroot3/sqroot3 to fix it.

4 0
2 years ago
YO PLEASE HELP MY TEST IS ALMOST DUE!!!!!!!!
Inga [223]

Answer:

3: x=-1

4: x=4

5: 0=0 infinite solutions

6: x=-3

3 0
3 years ago
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