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Vinvika [58]
3 years ago
8

Solve -8x < 2. A- B C D

Mathematics
1 answer:
Anarel [89]3 years ago
4 0

-8x <2

-8x÷2=4x

4× > 1

hope it helps

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List 2/3 , 7/12 , and 3/4 from greatest to least thanks .
nikklg [1K]
For this question you should know that:
2/3 = 8/12 
and 
3/4 = 9/12 
so now you can say:
7/12 < 8/12 < 9/12 
so: 
7/12 < 2/3 < 3/4 :)))
i hope this is helpful
have a nice day 
7 0
3 years ago
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Solve this quadratic form 3x-2x²=7 this is grade 9​
Pani-rosa [81]

Answer:

x

=

3

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i

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47

4

x=3±i474

Step-by-step explanation:

7 0
4 years ago
A line passes through (4,5,) which points is also on the line
goldfiish [28.3K]

Answer:yeet

Step-by-step explanation:

5 0
3 years ago
On a coordinate plane, a dashed straight line has a negative slope and goes through (0, 2) and (4, 0). Everything below and to t
yanalaym [24]

Answer:

Option C.

Step-by-step explanation:

A dashed straight line has a negative slope and goes through (0, 2) and (4, 0).

The given inequality is

y

We need find the point which is a solution to the given linear inequality.

Check the given inequality for point (2, 3).

3

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This statement is false. Option 1 is incorrect.

Check the given inequality for point (2, 1).

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Check the given inequality for point (3, -2).

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4 0
3 years ago
What is the difference between bernoulii equation and riccati equation
melisa1 [442]
The Bernoulli equation is almost identical to the standard linear ODE.

y'=P(x)y+Q(x)y^n

Compare to the basic linear ODE,

y'=P(x)y+Q(x)

Meanwhile, the Riccati equation takes the form

y'=P(x)+Q(x)y+R(x)y^2

which in special cases is of Bernoulli type if P(x)=0, and linear if R(x)=0. But in general each type takes a different method to solve. From now on, I'll abbreviate the coefficient functions as P,Q,R for brevity.

For Bernoulli equations, the standard approach is to write

y'=Py+Qy^n
y^{-n}y'=Py^{1-n}+Q

and substitute v=y^{1-n}. This makes v'=(1-n)y^{-n}y', so the ODE is rewritten as

\dfrac1{1-n}v'=Pv+Q

and the equation is now linear in v.

The Riccati equation, on the other hand, requires a different substitution. Set v=Ry, so that v'=R'y+Ry'=R'\dfrac vR+Ry'. Then you have

y'=P+Qy+Ry^2
\dfrac{v'-R'\dfrac vR}R=P+Q\dfrac vR+R\dfrac{v^2}{R^2}
v'=PR+\left(Q+\dfrac{R'}R\right)v+v^2

Next, setting v=\dfrac{u'}u, so that v'=\dfrac{uu''-(u')^2}{u^2}, allows you to write this as a linear second-order equation. You have

\dfrac{uu''-(u')^2}{u^2}=PR+\left(Q+\dfrac{R'}R\right)\dfrac{u'}u+\dfrac{(u')^2}{u^2}
u''-\left(Q+\dfrac{R'}R\right)u'+PRu=0
u''+Su'+Tu=0

where S=-\left(Q+\dfrac{R'}R\right) and T=PR.
3 0
3 years ago
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