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g100num [7]
3 years ago
6

Witch fraction is equivalent to 1/3

Mathematics
1 answer:
natulia [17]3 years ago
5 0

Answer:

i dont know child but hello

Step-by-step e

Step-by-step explanation:

oooooooooooooooooooof



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Find the slope of a line traveling through the points (8, -3) and (2, 6)(8,−3)and(2,6).
Sauron [17]

Answer:

not sure if this is what youre looking for but here

y=-3/2x+9

Step-by-step explanation:

6 0
3 years ago
If runners in a long distance race were to run straight from the starting line to the finish line they would run 13 kilometers.
tia_tia [17]

5^2 +x^2 = 13^2

25 + x^2 = 169

x^2 = 144

x = sqrt(144) = 12

 they run west for 12 KM

 12+5 = 17 total km

7 0
3 years ago
Can someone please help me im being timed ‼️
Ad libitum [116K]
125 because it’s symmetrical between both
3 0
3 years ago
Read 2 more answers
Question 10 Multiple Choice Worth 4 points)
Lady_Fox [76]
A is your answer bc the -2 directly affects the x value and since its negative it will move left
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
4 years ago
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