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finlep [7]
3 years ago
14

Pre cacl pleaseeeee help. the options are in the attachments as well

Mathematics
1 answer:
ivanzaharov [21]3 years ago
3 0

Answer:

The values of x so that y = \csc x have vertical asymptotes are -2\pi, -\pi, 0, \pi, 2\pi.

Step-by-step explanation:

The function cosecant is the reciprocal of the function sine and vertical asymptotes are located at values of x so that function cosecant becomes undefined, that is, when function sine is zero, whose periodicity is \pi. Then, the  vertical asymptotes associated with function cosecant are located in the values of x of the form:

x = 0\pm \pi\cdot i, \forall \,i\in \mathbb{N}_{O}

In other words, the values of x so that y = \csc x have vertical asymptotes are -2\pi, -\pi, 0, \pi, 2\pi.

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In a set of 3 numbers the first is a positive integer, the second is 3 more than the first and the 3rd is a square of the second
Molodets [167]

Given:

In a set of 3 numbers the first is a positive integer, the second is 3 more than the first and the 3rd is a square of the second.

To find:

The equation for the given situation if the sum of the numbers is 77.

Solution:

a. Let the first number in the set is x.

The second is 3 more than the first. So, the second number is (x+3).

The 3rd number is a square of the second. So, the third number is (x+3)^2.

Therefore, the first, second and third numbers are x,(x+3),(x+3)^2 respectively.

b. The sum of the numbers is 77.

First number + Second number + Third number = 77

c. So, the equation in terms of x is:

x+(x+3)+(x+3)^2=77

Therefore, the required equation is x+(x+3)+(x+3)^2=77.

d. On simplification, we get

x+x+3+x^2+6x+9=77                   [\because (a+b)^2=a^2+2ab+b^2]

x^2+(6x+x+x)+(3+9)=77

x^2+8x+12=77

Subtract 77 from both sides.

x^2+8x+12-77=77-77

x^2+8x-65=0

Therefore, the simplified form of the required equation is x^2+8x-65=0.

6 0
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Answer:  

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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