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timurjin [86]
3 years ago
7

Can someone help? i really don’t understand. question is in the picture

Mathematics
1 answer:
finlep [7]3 years ago
8 0

Answer:

1,3

Step-by-step explanation: The slope of this graph is 1,3

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Identify three solutions of the graph shown. Please help
barxatty [35]
From the graph, we can see that the graph crosses the x-axis at the point (1.5, 0) the graph also passes through point (1, 1) and the graph crosses the y-axis at the point (0, 3).

Therefore, points (1.5, 0), (1, 1) and (0, 3) are some of the solutions of the graph.
5 0
3 years ago
Which of the following segments is a radius of OK?<br> ОА. АМТ<br> ОВ. Вт<br> Ос. AMR<br> O D. KM
FrozenT [24]

Answer:

A) TA

Step-by-step explanation:

A radius goes from the center of a circle to the outside edge.  

RP connects two points on the outside edge of the circle without going through the center.  This is not a radius.

TR connects two opposite points on the outside edge of the circle while going through the center.  This is a diameter, not a radius.

GP connects two opposite points on the outside edge of the circle while going through the center.  This is a diameter, not a radius.

TA connects the center of the circle to the outside edge.  This is a radius.

8 0
3 years ago
Which is greater 3 miles or 5,000 yards? How much greater?
Brilliant_brown [7]
5,000 yards is greater
3 0
3 years ago
Read 2 more answers
What is the length of LINE GH?<br> And please explain why thank you:
lesya692 [45]

co-ordinates of point G are (1,2)

co-ordinates of point H are (3,-3)

so, by distance formula :

\sqrt{( {x1 - x1})^{2} + ( {y1 - y2})^{2}  }  \\  =  \sqrt{( {1- 3})^{2} + ( {2 - ( - 3)})^{2}  }  \\   \sqrt{( { - 2})^{2} + ( {2  + 3})^{2}  }  \\  \sqrt{( { - 2})^{2} + ( {5})^{2}  }  \\   \sqrt{4 + 25}  \\  \sqrt{29}

so, it is root(29) or 5.38 units

7 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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