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tatiyna
2 years ago
12

What is the value of -3r+ 8 when r=4

Mathematics
1 answer:
denis23 [38]2 years ago
6 0
Answer: -4.


Explanation: Because -3*4=-12. -12+8=-4.
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Find the ordered pair to represent a T in the equation t in the equation t=1/2u-v if u=(-1,4) and v=(3,-2)
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If 0 = 6n -36 what is the value of n -5
juin [17]
0=6n- 36

6n-36

6 x-5-36

35-36

n=-1 
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8 0
3 years ago
ABCD is a quadrilateral inscribed in a circle, as shown below:
german

Answer:

  • A. (x + 16) + (6x − 4) = 180

Step-by-step explanation:

Inscribed quadrilateral has opposite angles supplementary.

<u>So</u>

  • m∠A + m∠C = 180°

or

  • m∠B + m∠D = 180°.

<u>Since</u>

  • m∠A = x + 16,
  • m∠B = x,
  • m∠C = 6x - 4,
  • m∠D = 2x + 16

we can use either pair of angle measures to work out the value of x and then find the value of each angle.

<u>We can verify the first option is the only correct one.</u>

  • m∠A + m∠C = 180°
  • (x + 16) + (6x − 4) = 180
3 0
3 years ago
Please help me with geometry
mariarad [96]

Answer:

x = 8

Step-by-step explanation:

Set 2x + 3 = 3x - 5

Solve for x

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3 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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