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ANEK [815]
2 years ago
10

Which equations transform the parent secant graph to the parent cosecant graph?

Mathematics
2 answers:
Gnesinka [82]2 years ago
8 0

The answer is actually y = sec(x - pi/2)

jenyasd209 [6]2 years ago
7 0

Answer:

B

Step-by-step explanation:

y=sec(x+pi/2)

Took the test on e2020

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Pablo’s goal is to walk a total of 1034
Free_Kalibri [48]

Answer:

c

Step-by-step explanation:

because you add up all the days then you get the total

4 0
2 years ago
HURRY TEACHER DIDN'T PUT ON STUDY GUIDE
Andru [333]

1 step: Place the point of the compass on point M and draw an arc, making sure the width of the compass opening is greater than 1/2MN.

2 step: Place the point of the compass on point N and draw an arc that has the same radius as previous one.

3 step: Connect two points of intersection of drawn arcs. Obtained line will be bisector.

Answer: option D.

5 0
3 years ago
Read 2 more answers
10 x 10 - 4 divided by 7
TEA [102]

Answer:

13.7142857143

Step-by-step explanation:

If you need to round then do so, or i will comment the rounded version if needed.

3 0
2 years ago
Which of the following expressions correctly uses the properties of summations to represent
madam [21]

Answer:

Option C is correct.

7 \cdot \sum_{i=1}^{18} i^2 + 9 \cdot 18

Step-by-step explanation:

Given the expression:  \sum_{i=1}^{18} (7i^2+9)

Using properties of summation:

  • \sum_{i=1}^{n} (a+bi) =\sum_{i=1}^{n} a + \sum_{i=1}^{n} bi
  • \sum_{i=1}^{n} a = an

Using properties of summation in the given expression we have;

\sum_{i=1}^{18} (7i^2+9)

= \sum_{i=1}^{18} 7i^2 + \sum_{i=1}^{18} 9

=7 \cdot \sum_{i=1}^{18} i^2 + 9 \cdot 18

Therefore, the following given expression uses the properties of summation to represents is, 7 \cdot \sum_{i=1}^{18} i^2 + 9 \cdot 18


8 0
3 years ago
Read 2 more answers
How do you evaluate sin(13π/12)?
Alekssandra [29.7K]

13pi/12 lies between pi and 2pi, which means sin(13pi/12) < 0

Recall the double angle identity,

sin^2(x) = (1 - cos(2x))/2

If we let x = 13pi/12, then

sin(13pi/12) = - sqrt[(1 - cos(13pi/6))/2]

where we took the negative square root because we expect a negative value.

Now, because cosine has a period of 2pi, we have

cos(13pi/6) = cos(2pi + pi/6) = cos(pi/6) = sqrt[3]/2

Then

sin(13pi/12) = - sqrt[(1 - sqrt[3]/2)/2]

sin(13pi/12) = - sqrt[2 - sqrt[3]]/2

7 0
3 years ago
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