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blagie [28]
3 years ago
11

In a circle, a 90° sector has area 16π ft2. What is the radius of the circle?

Mathematics
1 answer:
Veronika [31]3 years ago
7 0
Sector area  = 0.5 * r^2 * angle (in radians)

so:

16 pi = 0.5 * r^2 * (90*pi/180)

r^2 = 16 pi /(0.25 pi) = 64

r = 8


therefore it would be d
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The factored form of a quadratic equation is y=(2x+1)(x-5), and the standard form is y=2x²-9x-5. Which of the following statemen
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x1= -5 x2=0.5 y= -5

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Step-by-step explanation:

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3 years ago
Hi can you help me ? what is 0.07 ? ​
riadik2000 [5.3K]

Answer:

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Step-by-step explanation:

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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
Find csc0 and cos0 , where 0 is the angle shown in the figure. Give exact values, not decimal approximations.
solniwko [45]

Answer:

csc(θ) =  1.1662

cos(θ)  = 0.5145

Step-by-step explanation:

I attach the image of your question in the picture below

We know the values of the adjacent and opposite cathetus

tan(θ) = Opposite / Adjacent

tan(θ) = 5/3 = 1.666

θ = tan^-1 (5/3)

θ = 59.0362 °

Now that we have the angle, we can easily find the value of the expressions needed

csc(θ) = 1 / sin(θ)  = 1/ (0.85749) = 1.1662

cos(θ)  = 0.5145

8 0
4 years ago
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