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2 years ago

What is the RANGE of the graph

Mathematics

2 answers:

Aleksandr [31]2 years ago

5 0

R: (-4,∞)

Lowest value of y is -4 and the graph goes on upwards continuously. The equation of the function is x^2 - 4

umka2103 [35]2 years ago

5 0

-4 to infinite because the bottom goes to -4 and the arrows don’t stop

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steposvetlana [31]

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2 years ago

a bicycle advertised all mountain bike priced at 1/3 discount what is the sales price if the original price is 327

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The discount sales price would be $218

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2 years ago

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Alexxx [7]

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Given this, there can be a relationship between the ideological group and the support of the moderate consideration act.
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brainly.com/question/18064632

#SPJ4

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1 year ago

Will mark brainliest

Varvara68 [4.7K]

Answer: I don’t think u gave enough info

Step-by-step explanation:

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2 years ago

Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.

Ann [662]

Note that if ${x_2}'=x_1$ , then ${x_2}''={x_1}'$ , and so we can collapse the system of ODEs into a linear ODE:

${x_2}''=3{x_2}'-x_2+e^t$
${x_2}''-3{x_2}'+x_2=e^t$

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

$r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0$

so that the characteristic solution is

${x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}$

Now let's suppose the particular solution is ${x_2}_p=ae^t$ . Then

${x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t$

and so

$ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1$

Thus the general solution for $x_2$ is

$x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t$

and you can find the solution $x_1$ by simply differentiating $x_2$ .

7 0

3 years ago