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seropon [69]
3 years ago
11

Find the solutions of each equation on the interval [0, 2π)​

Mathematics
1 answer:
Schach [20]3 years ago
4 0

Answer/Step-by-step explanation:

Factor this. GCF=2

2(2cos2x+cos x-1)=0

2(2cos x -1)(cos x+1)=0 Set each factor equal to zero and solve

2cos x -1=0 Add 1 to both sides

2cos x =1 Divide by 2

cos x =1/2

x=π/3

x=5π/3

cos x+1=0 Subtract 1 from both sides

cos x=-1

<u><em>x = π</em></u>

<u><em></em></u>

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In the diagram shown, AE 1/3AC and ED ll CB. Explain why df must be one-fourth length of CD
Blizzard [7]

The value of the length DF is 1/4 of that of CD because of the Angle Bisector Theorem for triangles.

<h3>What is the Angle Bisector Theorem?</h3>

The Angel Bisector Theorem states that if a ray bisects an angle of a triangle, it dissects the side of the triangle that is opposite to the dissected angle into parts that are proportionate to the adjacent sides.

Note that the angle that has been dissected here is ∠CBD. Thus,

\frac{DF}{CF} = \frac{BD}{BC}

See the link below for more about Angle Bisector Theorem of the Triangle:

brainly.com/question/21475560

8 0
2 years ago
5 inches is what fraction of a foot
ipn [44]

Answer:

5/12

Step-by-step explanation:

3 0
2 years ago
Choose whether the expression is a result of multiplying
vladimir1956 [14]

Answer:

neither

Step-by-step explanation:

Take out 5 as a common factor. It will be easier to look at.

5(5c^2 + 11c + 6)

5(5C +6 )(c + 1 )

Now you can put the 5 inside.

(25c + 30)(c + 1) is one answer.

(5c + 6)(5c + 5) is another.

The answer is multiplying binomials. There is nothing that that is squared and the answers are not conjugates. They are two binomials multiplied together.

8 0
2 years ago
What is 1 plus 2 plus 3.... plus 200?
Aneli [31]

Answer:

20100

Step-by-step explanation:

There are 200 numbers in this addition problem. To add this, we can do 1 + 200, 2 + 199, 3 + 198 and so on. All of these pairs have a sum of 201 and since there are 200 / 2 = 100 pairs the answer is 201 * 100 = 20100.

6 0
3 years ago
Read 2 more answers
AYUDA CON ESTO!!! ALGUIEN PORFAVOR
Gre4nikov [31]

Answer:

Problem 1)  frequency:  160 heartbeats per minute, period= 0.00625 minutes (or 0.375 seconds)

Problem 2) Runner B has the smallest period

Problem 3) The sound propagates faster via a solid than via air, then the sound of the train will arrive faster via the rails.

Step-by-step explanation:

The frequency of the football player is 160 heartbeats per minute.

The period is (using the equation you showed above):

Period = \frac{1}{frequency} = \frac{1}{160} \,minutes= 0.00625\,\,minutes = 0.375\,\,seconds

second problem:

Runner A does 200 loops in 60 minutes so his frequency is:

\frac{200}{60} = \frac{10}{3} \approx  3.33   loops per minute

then the period is: 0.3 minutes (does one loop in 0.3 minutes)

the other runner does 200 loops in 65 minutes, so his frequency is:

\frac{200}{65} = \frac{40}{13} \approx  3.08   loops per minute

then the period is:

\frac{13}{40} =0.325\,\,\,minutes

Therefore runner B has the smaller period

8 0
3 years ago
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