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-BARSIC- [3]
3 years ago
6

What are the solutions of the equation x^6-9x^3+8=0

Mathematics
1 answer:
Triss [41]3 years ago
4 0

USE CALCULATOR!!! EPLAINT OOOO

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Which equation(s) model inverse variation? Select all that apply.
Harman [31]

For this case, what you should know is that the equations that represent an inverse variation are those that could not form a straight line, for example.

We have then:

Equation 1:

pv = 13

Rewriting:

p = 13 / v

P and v are represent an inverse variation.

Equation 2:

z = (2 / x)

z and x are represent an inverse variation.

Answer:

equations represented inverse variation are:

pv = 13

z = (2 / x)

7 0
3 years ago
Use the laplace transform to solve the given initial-value problem. y' 5y = e4t, y(0) = 2
Basile [38]

The Laplace transform of the given initial-value problem

y' 5y = e^{4t}, y(0) = 2 is  mathematically given as

y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}

<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>

Generally, the equation for the problem is  mathematically given as

&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}

\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}

put $s=-s \Rightarrow a_{1}=\frac{17}{9}$

\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}

In conclusion, Taking inverse Laplace tranoform

L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\

y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}

Read more about Laplace tranoform

brainly.com/question/14487937

#SPJ4

6 0
1 year ago
The vertical shift, k, is the amount every
HACTEHA [7]

Answer: k= 4

Step-by-step equation

3 0
3 years ago
Read 2 more answers
Question part points submissions used consider the following. g(x)={((x**2-a**2)/(x-a) text(if ) x != a,4 text(if ) x =
EastWind [94]
For x\neq a, we have

\dfrac{x^2-a^2}{x-a}=\dfrac{(x-a)(x+a)}{x-a}=x+a

So for g(x) to be continuous at x=a, we require that the limit as x\to a is equal to 4.

\displaystyle\lim_{x\to a}g(x)=\lim_{x\to a}(x+a)=2a=4\implies a=2
4 0
3 years ago
Amari claims that a rectangle is sometimes a parallelogram but a parallelogram is always a
Maslowich

Answer:

The statement is false.

Step-by-step explanation:

A parallelogram is a figure of four sides, such that opposite sides are parallel

A rectangle is a four-sided figure such that all internal angles are 90°

Here, the statement is:

"A rectangle is sometimes a parallelogram but a parallelogram is always a

rectangle."

Here if we found a parallelogram that is not a rectangle, then that is enough to prove that the statement is false.

The counterexample is a rhombus, which is a parallelogram that has two internal angles smaller than 90° and two internal angles larger than 90°, then this parallelogram is not a rectangle, then the statement is false.

The correct statement would be:

"A parallelogram is sometimes a rectangle, but a rectangle is always a parallelogram"

6 0
2 years ago
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