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kipiarov [429]
3 years ago
13

A circular plate has circumference of 26.4 inches. What is the area of this​ plate? Use 3.14 for pi.

Mathematics
1 answer:
mafiozo [28]3 years ago
3 0
I believe the answer is 55.49 inches
using the equation C^2/(4*3.14)
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i believe its a

Step-by-step explanation:

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2 years ago
Graph ƒ(x) = 2x - 3.
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Ok so your gonna go on the graph and your gonna go to -3 and make a dot then from that point you go up two times and over to the right once and that's your slope
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Helllpppppppppp please someone.?????
Liula [17]

Answer:

(-5,5)

Step-by-step explanation:

Multiply the first equation by -2,and multiply the second equation by 1.

−2(x+5y=20)

1(2x−7y=−45)

Becomes:

−2x−10y=−40

2x−7y=−45

Add these equations to eliminate x:

−17y=−85

Then solve−17y=−85for y:

(Divide both sides by -17)

y=5

x+5y=20

Substitute  5 for y in x+5y=20:

x+(5)(5)=20

x+25=20(Simplify both sides of the equation)

x+25+−25=20+−25(Add -25 to both sides)

x=-5

6 0
3 years ago
Sarah and Jake timed themselves while running. Who ran at a faster speed?
RUDIKE [14]

Answer:

Jake

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4 0
3 years ago
Use the substitution of x=e^{t} to transform the given Cauchy-Euler differential equation to a differential equation with consta
kherson [118]

By the chain rule,

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\implies\dfrac{\mathrm dy}{\mathrm dt}=x\dfrac{\mathrm dy}{\mathrm dx}

which follows from x=e^t\implies t=\ln x\implies\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x.

\dfrac{\mathrm dy}{\mathrm dt} is then a function of x; denote this function by f(x). Then by the product rule,

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dt}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac1x\dfrac{\mathrm df}{\mathrm dx}

and by the chain rule,

\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dt^2}

so that

\dfrac{\mathrm d^2y}{\mathrm dt^2}-\dfrac{\mathrm dy}{\mathrm dt}=x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}

Then the ODE in terms of t is

\dfrac{\mathrm d^2y}{\mathrm dt^2}+8\dfrac{\mathrm dy}{\mathrm dt}-20y=0

The characteristic equation

r^2+8r-20=(r+10)(r-2)=0

has two roots at r=-10 and r=2, so the characteristic solution is

y_c(t)=C_1e^{-10t}+C_2e^{2t}

Solving in terms of x gives

y_c(x)=C_1e^{-10\ln x}+C_2e^{2\ln x}\implies\boxed{y_c(x)=C_1x^{-10}+C_2x^2}

4 0
3 years ago
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