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Aleks [24]
3 years ago
12

The vertical _____ of a function secant are determined by the points that are not in the domain.

Mathematics
2 answers:
shusha [124]3 years ago
7 0
The vertical asymptotes of a function secant are determined by the points that are not in domain.
Thank you.
luda_lava [24]3 years ago
5 0

Answer:

The vertical <u>asymptote</u> of a function secant are determined by the points that are not in the domain.

Step-by-step explanation:

The domain of a function is the set of x values for which the function is defined.

Secant function is not defined at x=\frac{\pi}{2}+n\pi

It means we cannot include these points in the domain.

At these points, we must have a vertical line which do not touch the graph. These lines are called "Vertical asymptotes"

Vertical asymptotes are not included in the domain of the function.

Hence, the correct word should be "Asymptote"

The vertical <u>asymptote</u> of a function secant are determined by the points that are not in the domain.

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Whats the answer to this ?
8090 [49]

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Step-by-step explanation:

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2 years ago
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uysha [10]

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8 0
3 years ago
Read 2 more answers
What is the y-coordinate of the point that divides the directed
e-lub [12.9K]

Answer:

(D)5

Step-by-step explanation:

Given the point J(-3,1) and K(8,11).

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P(x,y)=\left(\dfrac{mx_2+nx_1}{m+n} ,\dfrac{my_2+ny_1}{m+n}\right)

In the given case:

(x_1,y_1)=(-3,1), (x_2,y_2)=(8,11), m:n=2:3

Since we are to determine the y-coordinate of the point that divides JK into a ratio of 2:3, we have:

\dfrac{my_2+ny_1}{m+n}=\dfrac{2*11+3*1}{3+2}\\\\=\dfrac{22+3}{5}\\\\=\dfrac{25}{5}\\\\=5

The y-coordinate of the point that divides the directed  line segment from J to K into a ratio of 2:3 is 5.

The correct option is D.

3 0
3 years ago
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