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sasho [114]
3 years ago
5

Which of the functions below have period 2pi? Check all that apply. A. y = tan x B. y = sec x C. y = cot x D. y = csc x

Mathematics
2 answers:
swat323 years ago
8 0
The correct answers are y=sec x, and y=csc x
Colt1911 [192]3 years ago
4 0

Answer:

Option B and D are correct that is y=secx and  y=cscx  respectively.

Step-by-step explanation:

We have been given four functions:

CaseA:

y=tanx

\text{Period oftanx is} \pi

Hence, A is correct.

CaseB:

y=secx

\text{Period of secx is} 2\pi

Hence, B is correct.

CaseC:

y=cotx

\text{Period of cotx is} \pi

Hence, C is not correct.

CaseD:

y=cscx

\text{Period of cscx is} 2\pi

Hence, D is correct.

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Line a is parallel to line b, m<2 = 4x+44 , And m<6=6x+36. Find the value of x
Savatey [412]

Answer:

x = 4

Step-by-step explanation:

<2 and <6 are equal because of the corresponding angles thm so:

4x + 44 = 6x + 36

4x - 6x = 36 - 44

-2x = -8

x = 4

3 0
3 years ago
Solve the equation<br><br> 1 . J + -2 = -22
olganol [36]

J + -2 = -22

Adding a negative number changes the equation to subtract ruin:

J - 2 = -22

Now add 2 to both sides to get the answer:

J = -20

7 0
3 years ago
Read 2 more answers
Three-fourths of p is 18
never [62]
<h3><u>The value of p is 24.</u></h3>

3/4p = 18

Multiply both sides by 4.

3p = 72

Divide both sides by 3.

p = 24


8 0
3 years ago
What kind of sequence is the pattern 15, 17, 20, 24, 29,
VikaD [51]

\implies {\blue {\boxed {\boxed {\purple {\sf { \: 15, 17, 20, 24, 29,35, 42}}}}}}

\sf \bf {\boxed {\mathbb {Step-by-step\:explanation:}}}

\: 15, 17, 20, 24, 29,35, 42

➺ \: 15 + 2 = 17

➺\: 17 + 3 = 20

➺\: 20 + 4 = 24

➺\: 24 + 5 = 29

➺\: 29 + 6 = 35

➺\: 35 + 7 = 42

\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{❦}}}}}

7 0
3 years ago
I need help with this pls
antoniya [11.8K]
I'll explain it simply for you

1st question
Of course you know phythagoras theorm
You even wrote it up there
It states that the sum of the square of the two sides of an equilateral triangle is equal to the square of the hypotenuse
{a}^{2}  +  {b}^{2}   =  {c}^{2}
Where C is the hypotenuse
*NOTE* :
HYPOTENUSE is the greatest side in a triangle!!
And that's where your mistake is!
So you should take the greatest side as C
So in Q3. 7, 24 and 26 are the given numbers
You'll make the smaller two numbers a and b and the greatest number C
Using the Formula you'll solve the left side first which is
{a}^{2}  +  {b}^{2}
Then the right side which is
{c}^{2}

And if both are equal then it is a right triangle otherwise it isn't!
Let
a=7
b=24
c=26


a^2 + b^2
7^2 + 24^2
49 + 576 = 625
GREAT, Now the right side

26^2 = 676
Since they aren't equal it isn't a right angled triangle...

Then let
a=7.5
b=10
c=12.5

7.5^2 + 10^2
= 56.25 + 100
= 156.25



12.5^2 = 156.25

They are EQUAL
Therefore it is a right triangle too

Hopefully I helped
7 0
3 years ago
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