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adell [148]
10 months ago
11

Given f(x)=e^-x^3 find the vertical and horizontal asymptotes

Mathematics
1 answer:
Aleonysh [2.5K]10 months ago
5 0

Given:

f\mleft(x\mright)=e^{-x^3}

To find the vertical and horizontal asymptotes:

The line x=L is a vertical asymptote of the function f(x) if the limit of the function at this point is infinite.

But, here there is no such point.

Thus, the function f(x) doesn't have a vertical asymptote.

The line y=L is a vertical asymptote of the function f(x) if the limit of the function (either left or right side) at this point is finite.

\begin{gathered} y=\lim _{x\rightarrow\infty}e^{-x^3} \\ =e^{-\infty} \\ y=0 \\ y=\lim _{x\rightarrow-\infty}e^{-x^3} \\ y=e^{\infty} \\ =\infty \end{gathered}

Thus, y = 0 is the horizontal asymptote for the given function.

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Cinco menos tres cuartos x ?
Arte-miy333 [17]

Answer:

4.25

Step-by-step explanation:

5-3/4=4.25

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3 0
3 years ago
Write the equation in slope-intercept form of the line that has a slope of -7/8 and contains the point (-4, 9).
lubasha [3.4K]

Answer:

The equation of the line is y = -7/8x + 11/2

Step-by-step explanation:

Since we have the slope and a point to start, we can use point-slope form to find the equation.

y - y1 = m(x - x1)

Use the slope for m and the point at (x1, y1). Then solve for y.

y - 9 = -7/8(x + 4)

y - 9 = -7/8x - 7/2

y = -7/8x + 11/2

6 0
3 years ago
Find the values of the sine, cosine, and tangent for ZA.<br><br> (TOP OF TRIANGLE IS (A))
Sloan [31]

\bigstar\:{\underline{\sf{In\:right\:angled\:triangle\:ABC\::}}}\\\\

  • AC = 7 m
  • BC = 4 m

⠀⠀⠀

\bf{\dag}\:{\underline{\frak{By\:using\:Pythagoras\: Theorem,}}}\\\\

\star\:{\underline{\boxed{\frak{\purple{(Hypotenus)^2 = (Perpendicular)^2 + (Base)^2}}}}}\\\\\\ :\implies\sf (AB)^2 = (AC)^2 + (BC)^2\\\\\\ :\implies\sf (AB)^2 = (AB)^2 = (7)^2 = (4)^2\\\\\\ :\implies\sf (AB)^2 = 49 + 16\\\\\\ :\implies\sf (AB)^2 = 65\\\\\\ :\implies{\underline{\boxed{\pmb{\frak{AB = \sqrt{65}}}}}}\:\bigstar\\\\

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━

☆ Now Let's find value of sin A, cos A and tan A,

⠀⠀⠀

  • sin A = Perpendicular/Hypotenus = \sf \dfrac{4}{\sqrt{65}} \times \dfrac{\sqrt{65}}{\sqrt{65}} = \pink{\dfrac{4 \sqrt{65}}{65}}

⠀⠀⠀

  • cos A = Base/Hypotenus = \sf \dfrac{7}{\sqrt{65}} \times \dfrac{\sqrt{65}}{\sqrt{65}} = \pink{\dfrac{7 \sqrt{65}}{65}}

⠀⠀⠀

  • tan A = Perpendicular/Base = {\sf{\pink{\dfrac{4}{7}}}}

⠀⠀⠀

\therefore\:{\underline{\sf{Hence,\: {\pmb{Option\:A)}}\:{\sf{is\:correct}}.}}}

4 0
2 years ago
Evaluate the expression when c=4 and d=48. d-6c​
Rama09 [41]

Answer:

24

Step-by-step explanation:

since c=4 and d=48, we can re-write this equation

48-6 x 4

48 - 24

24

5 0
2 years ago
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