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blondinia [14]
3 years ago
5

Which is an asymptote of the graph of the function y=tan(3/4x)?

Mathematics
2 answers:
abruzzese [7]3 years ago
7 0

ANSWER

x =  -  \frac{2\pi}{3}

EXPLANATION

The given function has equation

y =  \tan( \frac{3}{4}x)

This can be rewritten as

y =  \frac{ \sin( \frac{3}{4}x ) } { \cos(\frac{3}{4}x )}

The asymptote occurs at:

\cos(\frac{3}{4}x ) = 0

This implies that,

\frac{3}{4} x =  \frac{ \pi}{2}

x =  \frac{ \pi}{2}  \times  \frac{4}{3}

x =  \frac{2\pi}{3}

Or

\frac{3}{4} x =  -  \frac{ \pi}{2}

x =  \frac{ -  \pi}{2}  \times  \frac{4}{3}

x =   - \frac{2\pi}{3}

The second choice is correct.

MArishka [77]3 years ago
6 0

Answer:

B) x = \frac{2\pi }{3}

Step-by-step explanation:

this is the correct answer on ed-genuity, hope this helps! :)

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4 years ago
For the following pair of functions, find (f+g)(x) and (f-g)(x).
ch4aika [34]

Given:

The functions are

f(x)=4x^2+7x-5

g(x)=-9x^2+4x-13

To find:

The functions (f+g)(x) and (f-g)(x).

Solution:

We know that,

(f+g)(x)=f(x)+g(x)

(f+g)(x)=4x^2+7x-5-9x^2+4x-13

(f+g)(x)=(4x^2-9x^2)+(7x+4x)+(-5-13)

(f+g)(x)=-5x^2+11x-18

And,

(f-g)(x)=f(x)-g(x)

(f-g)(x)=(4x^2+7x-5)-(-9x^2+4x-13)

(f+g)(x)=4x^2+7x-5+9x^2-4x+13

(f+g)(x)=(4x^2+9x^2)+(7x-4x)+(-5+13)

(f-g)(x)=13x^2+3x+8

Therefore, the required functions are (f+g)(x)=-5x^2+11x-18

and (f-g)(x)=13x^2+3x+8.

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13x - 6 = 8x + 19 solve for x
klio [65]
13x=8x+25
5x=25
X=5


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