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Sonja [21]
2 years ago
10

Any body wana play imvu.........​

Mathematics
2 answers:
Igoryamba2 years ago
7 0
Dfkcghkbftff is that
natita [175]2 years ago
5 0

Answer:

sure...

Step-by-step explanation:

brainliest plz

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For the polynomial below, find F(-3)<br><br> F(-3)= -4x^2+3x-1
Elodia [21]
f(-3)=-4\cdot(-3)^2+3\cdot(-3)-1=-4\cdot9-9-1=-36-10=-46
4 0
3 years ago
How to find equation hyperbola with asymptotes, center and transverse?
lapo4ka [179]
The equation for a hyperbola is (x-h)/a - (y-k)/b = 1
Or (y-k)/a - (x-h)/b = 1
h represents the x value of the coordinate
k value represents the y value of the coordinate 
together they represent a point, which is the center
So (h,k) is (x,y)

The asymptote is y-k = +/- b/a (x-h)
The transverse is the line that goes through the hyperbola. 
6 0
3 years ago
Find the value of x for the given triangle
11Alexandr11 [23.1K]

Answer:

5

Step-by-step explanation:

6 0
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
(-8a^2 - 1)(3a^2 - 5)
natali 33 [55]

Answer:

13a² + 5

Step-by-step explanation:  

Not 100% but 90% sure. Work on Picture.

4 0
3 years ago
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