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3 years ago

If csc 0 = 2, find sin 0

Mathematics

1 answer:

Slav-nsk [51]3 years ago

3 0

Answer:

1/2 (Please read explanation to educate yourself)

Step-by-step explanation:

csc = hyp/opp

sin is the inverse of csc --> sin = opp/hyp

so for csc you have csc(0) = 2/1 and since sin is the inverse then sin(0) = 1/2

Very basic trigonometry and it is the fundamentals of every higher math so learn it well.

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What is the principal for simple interest interest is 3 12.50 rate is 25% for 25 years

lilavasa [31]

$312.50 = p x 0.25 x 25
divided by 25
12.50 = p x 0.25
divide by 0.25
50 = principal
Just ask if anything is unclear

6 0

3 years ago

Find the 8th term of the geometric sequence 3, 9, 27, ...

miv72 [106K]

Answer:

8th term- 6,561

Step-by-step explanation:

Each term in the sequence is being multiplied by 3. For example, 3x3=9 and 9x3=27. Now, we need to continue multiplying numbers, so we can list them all here, still starting with 3.

1&2) 3x3=9

3) 9x3=27

4) 27x3=81

5) 81x3=243

6) 243x3=729

7) 729x3=2,187

8) 2,187x3=6,561

The 8th term of the geometric sequence is 6,561.

Hope this helps and have a great day! ^^

3 0

3 years ago

Last question need to be answered

shepuryov [24]

The answer would be 110 degrees.

4 0

4 years ago

Read 2 more answers

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$