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Sergio039 [100]
3 years ago
14

What is the slope of a line perpendicular to line B

Mathematics
1 answer:
NemiM [27]3 years ago
3 0

Answer:


Step-by-step explanation:

Where is line B?

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natita [175]

Answer:

Y= -6x + 1

Step-by-step explanation:

Y=mx + b

M is your slope

B is your y intercept

5 0
3 years ago
A jewelry store offers its own brand of customizable charm bracelets. You are able to choose from 9
Elis [28]

Answer:

By the way that it is worded, I believe that you are only allowed to put 1 charm on the bracelet, so there are 9 ways to pick a charm.

Hope this helps(and also if you have a more refined wording please put it in the comments)!

4 0
3 years ago
I have 183 feet of fencing. The length of the court is 32 feet more than the width. One of the shorter sides will be 5 feet shor
Aliun [14]
Okay so the total is 183. So we can say 183 = 
Let's call the width (the longer side) x.
We know that we have x, x + 32, and x - 5.
We can say 183 = x + (x + 32) + (x - 5).
Let's solve for x.
x = 52. 
So we know the length is 84 feet, the shorter side has 47 feet, and the longer side has 52 feet.
7 0
2 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
6.
disa [49]

Answer:

1/3

goes up one over three

rise over run

3 0
2 years ago
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