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3 years ago

What are the vertical and horizontal asymptotes for the function f(x)= x^2+x-6/x^3-1?

Mathematics

2 answers:

weeeeeb [17]3 years ago

8 0

Answer:

Option B

Step-by-step explanation:

Given

f(x)= (x^2+x-2)/(x^3-1)

For vertical asymptotes we have to put the denominator of the function equal to zero:

x^3-1=0

x^3=1

So,

x=1

As the degree of denominator is greater than the degree of numerator, there will be only horizontal asymptote which will be y=0 because all the y values will be dragged down to the x-axis.

So the horizontal asymptote: y=0

And Vertical asymptote: x =1

telo118 [61]3 years ago

4 0

Answer:

Step-by-step explanation:

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If JKLM is a rhombus, MK = 30, NL = 13, and mZMKL = 41°, find each measure.

oksian1 [2.3K]

Answer:

$NK = 15$

$JL = 26$

$KL = 19.85$

$\angle JKM =49$

$\angle JML =41$

$\angle MLK = 90$

$\angle MNL =90$

$\angle KJL =41$

Step-by-step explanation:

Given

$MK = 30$

$NL = 13$

$\angle MKL = 41$

Solving (a): NK

MK is a diagonal and NK is half of the diagonal. So:

$NK = \frac{1}{2} * MK$

$NK = \frac{1}{2} * 30$

$NK = 15$

Solving (b): JL

JL is a diagonal, and it is twice of NL.

$JL = 2 * NL$

$JL = 2 * 13$

$JL = 26$

Solving (c): KL

To solve for KL, we consider triangle KNL where:

$\angle KNL = 90$

and

$KL^2 = NL^2 + NK^2$

$KL^2 = 13^2 + 15^2$

$KL^2 = 394$

$KL = \sqrt{394$

$KL = 19.85$

Solving (d - h):

To do this, we consider triangle JKN

$\angle KNL = \angle LNM = \angle MNJ = \angle JNK = 90$ -- diagonals bisect one another at right angle

Alternate interior angles are equal. So:

$\angle MKL = \angle KMJ = \angle KJL = \angle JLM = 41$

Similarly:

$\angle MKJ = \angle KML = \angle MJL = \angle JLK = 90 - 41$

$\angle MKJ = \angle KML = \angle MJL = \angle JLK = 49$

So:

$\angle JKM =49$

$\angle JML =41$

$\angle MLK = \angle MLJ + \angle JLK$

$\angle MLK = 49 + 41$

$\angle MLK = 90$

$\angle MNL =90$

$\angle KJL =41$

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irakobra [83]

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deff fn [24]

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Henry and reiko both used 1 yard of ribbon to make bows. write two different fractions to show that Henry and reiko use the same

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3 years ago

PLEASE ANSWER QUICKLY

Luba_88 [7]

29-11=18
18²+18²=p²
324+324=p²
p²=648
p=18√2
That's your answer.

6 0

3 years ago

Read 2 more answers