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KIM [24]
3 years ago
10

Is f(x)=5x^3/7 an exponential function?

Mathematics
2 answers:
Temka [501]3 years ago
6 0

Step-by-step explanation:

yes it is an exponential function.

Murljashka [212]3 years ago
3 0

Answer:

Yes

Step-by-step explanation:

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Please help i don't know this at all
melamori03 [73]

Answer:

simplify the expressions

Step-by-step explanation:

you just have to simplify all the different expression and then you can see the ones that are the same

6 0
4 years ago
A SSS<br>B AAA<br>C SSA<br>D None of the choices are correct ​
Mrac [35]

Answer:

B

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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
4 years ago
Please i need this quick , it’s so confusing
Dimas [21]
I think its A : 25 B : 81 C: 49
3 0
3 years ago
Refer to Expression 1 and Expression 2.
DanielleElmas [232]
The answer would be -345 because you multiply first, then add, and last subtract.
5 0
3 years ago
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