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What is a chcade in f 1 moter raceing?

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Anonymous

 ∙ 10y ago
Updated: 8/21/2019

Perhaps you mean "chicane," an artificial narrowing or turn on a road or auto-racing course.

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Wiki User

 ∙ 10y ago

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Related Questions

Install blower motor 2004 Ford F-150?

instalation of f150 blower moter which way does it mount?

You transplanted a 360 into a 74 CJ5 and the carb leaks due to the slant of the motor How do you fix this?

If your carb leaks because the moter is that slanted, then your moter mounts are totally f--- upped! Some CJ5`s came with v8`s as an option so you should be able to find the correct mounts in a junk yard. They are very common.

Why is the sum of the reciprocals of all of the divisors of a perfect number equal to 2?

Suppose N is a perfect number. Then N cannot be a square number and so N has an even number of factors.Suppose the factors are f(1) =1, f(2), f(3), ... , f(k-1), f(k)=N.Furthermore f(r) * f(k+1-r) = N for r = 1, 2, ... k so that f(r) = N/f(k+1-r)which implies that 1/f(r) = f(k+1-r)/NThen 1/f(1) + 1/(f(2) + ... + 1/f(k)= f(k)/N + f(k-1)/N + ... + f(1)/N= [f(k) + f(k-1) + ... + f(1)] / N= 2N/N since, by definition, [f(k) + f(k-1) + ... + f(1)] = 2N

Can you put a 1995 f-250 e40d transmission in a 1990 f-250 with a e40d tran they both have a 5.8 moter and they do look the same?

yes if they are both the same size front to back ans side to side and make shur all the conectors and so on are the same

F(x)=2x+3,a=-1 find 1/f. And graph f and 1/f?

Given the function f(x) = 2x + 3 and a = -1, we can find f(a) as follows: f(a) = 2(-1) + 3 f(a) = -2 + 3 f(a) = 1 So, f(a) = 1. To graph f(x) and 1/f(x), we can plot several points and connect them to visualize the functions. Here are some points for f(x): For f(x): When x = -2, f(x) = 2(-2) + 3 = -1 When x = -1, f(x) = 2(-1) + 3 = 1 When x = 0, f(x) = 2(0) + 3 = 3 When x = 1, f(x) = 2(1) + 3 = 5 When x = 2, f(x) = 2(2) + 3 = 7 Now, to find 1/f(x), we take the reciprocal of each f(x) value: For 1/f(x): When x = -2, 1/f(x) = 1/(-1) = -1 When x = -1, 1/f(x) = 1/1 = 1 When x = 0, 1/f(x) = 1/3 ≈ 0.333 When x = 1, 1/f(x) = 1/5 ≈ 0.2 When x = 2, 1/f(x) = 1/7 ≈ 0.143 Now, we can plot these points and connect them to obtain the graphs of f(x) and 1/f(x).

How To Solve A Function F(x) Has These Properties The Domain Of F Is The Set Of Natural Numbers F (1)1 F (x plus 1)f(x) plus 3x(x plus 1) plus 1 A. Determine F (2) F(3) F (4) F (5) F(6) THANKS!!!!!!?

The function (sequence generator) f(x) with x∈ℕ has been defined as a recursive function (sequence), with the initial value defined for some x, ie starting form some some natural number (in this case 1), the value of the function (sequence) is given (in this case f(1) = 1), and each successive value of the function (sequence) is defined in terms of the current value f(x+1) = f{x} + g(x) where g(x) is a function with g(x) = 3x(x + 1).f(x + 1) = f(x) + 3x(x + 1)f(1) = 1→ f(2) = f(1 + 1) = f(1) + (3×1)(1 + 1) = 1 + 3×2 = 1 + 6 = 7→ f(3) = f(2 + 1) = f(2) + (3×2)(2 + 1) = 7 + 6×3 = 7 + 18 = 25I'll let you evaluate the rest.Hint:f(4) = f(3 + 1) = f(3) + (3×3)(3 + 1)f(5) = f(4 + 1) = f(4) + ...f(6) = f(5 + 1) = f(5) + ...

How To Solve A Function F(x) Has These Properties The Domain Of F Is The Set Of Natural Numbers F (1)1 F (x plus 1)f(x) plus 3x(x plus 1) plus 1 Determine F (2) F(3) F (4) F (5) F(6) Describe The Func?

Given the properties of the function ( F(x) ), we can start by calculating ( F(1) = 1 ). Using the recursive relation ( F(x+1) = F(x) + 3x(x+1) + 1 ), we can compute the subsequent values: ( F(2) = F(1) + 3 \cdot 1 \cdot 2 + 1 = 1 + 6 + 1 = 8 ) ( F(3) = F(2) + 3 \cdot 2 \cdot 3 + 1 = 8 + 18 + 1 = 27 ) ( F(4) = F(3) + 3 \cdot 3 \cdot 4 + 1 = 27 + 36 + 1 = 64 ) ( F(5) = F(4) + 3 \cdot 4 \cdot 5 + 1 = 64 + 60 + 1 = 125 ) ( F(6) = F(5) + 3 \cdot 5 \cdot 6 + 1 = 125 + 90 + 1 = 216 ) The values are ( F(2) = 8 ), ( F(3) = 27 ), ( F(4) = 64 ), ( F(5) = 125 ), and ( F(6) = 216 ). The function exhibits a polynomial growth pattern, possibly resembling ( F(n) = n^3 ) as the outputs correspond to cubes of natural numbers.

A f 1 and 1 f a?

All For 1 and 1 For All

What is 3 over 6 and 1 over f?

3/6 + 1/f = 1/2 + 1/f = (f+2)/2f

When was Pierre De fermat's last theorem created?

PIERRE DE FERMAT's last Theorem. (x,y,z,n) belong ( N+ )^4.. n>2. (a) belong Z F is function of ( a.) F(a)=[a(a+1)/2]^2 F(0)=0 and F(-1)=0. Consider two equations F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) We have a string inference F(z)=F(x)+F(y) equivalent F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) infer F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) infer F(z-x-2)=F(x-x-2)+F(y-x-2) we see F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) give z=y and F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) give z=/=y. So F(z-x-1)=F(x-x-1)+F(y-x-1) don't infer F(z-x-2)=F(x-x-2)+F(y-x-2) So F(z)=F(x)+F(y) don't infer F(z-1)=F(x-1)+F(y-1) So F(z)=F(x)+F(y) is not equivalent F(z-1)=F(x-1)+F(y-1) So have two cases. [F(x)+F(y)] = F(z) and F(x-1)+F(y-1)]=/=F(z-1) or vice versa So [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). Or F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). We have F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. So x^3+y^3=/=z^3. n>2. .Similar. We have a string inference G(z)*F(z)=G(x)*F(x)+G(y)*F(y) equivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) we see G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) give z=y. and G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 infer G(x)>0. give z=/=y. So G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) don't infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) don't infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) is not equiivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So have two cases [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) and [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) or vice versa. So [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. Or G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] We have x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] So x^n+y^n=/=z^n Happy&Peace. Trần Tấn Cường.

Who can solve FLT?

Последнее Пьер де Ферма теоремы. (x,y,z,n) принадлежать( N+ )^4. n>2. (a) принадлежать Z F является функцией( a.) F(a)=[a(a+1)/2]^2 F(0)=0 и F(-1)=0. Рассмотрим два уравнения F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) непрерывный дедуктивного рассуждения F(z)=F(x)+F(y) эквивалент F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) выводить F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) выводить F(z-x-2)=F(x-x-2)+F(y-x-2) мы видим, F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) давать z=y и F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) давать z=/=y. так F(z-x-1)=F(x-x-1)+F(y-x-1) не выводить F(z-x-2)=F(x-x-2)+F(y-x-2) так F(z)=F(x)+F(y) не выводить F(z-1)=F(x-1)+F(y-1) так F(z)=F(x)+F(y) не эквивалентен F(z-1)=F(x-1)+F(y-1) Таким образом, возможны два случая. [F(x)+F(y)] = F(z) и F(x-1)+F(y-1)]=/=F(z-1) или наоборот так [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). или F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). у нас есть F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. так x^3+y^3=/=z^3. n>2. аналогичный непрерывный дедуктивного рассуждения G(z)*F(z)=G(x)*F(x)+G(y)*F(y) эквивалент G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) выводить G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) выводить G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) мы видим, G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) давать z=y. и G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 выводить G(x)>0. давать z=/=y. так G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y)не выводить G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) так G(z)*F(z)=G(x)*F(x)+G(y)*F(y) не выводить G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) так G(z)*F(z)=G(x)*F(x)+G(y)*F(y) не эквивалентен G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) Таким образом, возможны два случая. [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) и [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) или наоборот. так [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. или G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] у нас есть x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] так x^n+y^n=/=z^n Счастливые и мира. Trần Tấn Cường.

What is shortest solve about fermat?

To: trantancuong21@yahoo.com PIERRE DE FERMAT's last Theorem. (x,y,z,n) belong ( N+ )^4.. n>2. (a) belong Z F is function of ( a.) F(a)=[a(a+1)/2]^2 F(0)=0 and F(-1)=0. Consider two equations F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) We have a string inference F(z)=F(x)+F(y) equivalent F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) infer F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) infer F(z-x-2)=F(x-x-2)+F(y-x-2) we see F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) give z=y and F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) give z=/=y. So F(z-x-1)=F(x-x-1)+F(y-x-1) don't infer F(z-x-2)=F(x-x-2)+F(y-x-2) So F(z)=F(x)+F(y) don't infer F(z-1)=F(x-1)+F(y-1) So F(z)=F(x)+F(y) is not equivalent F(z-1)=F(x-1)+F(y-1) So have two cases. [F(x)+F(y)] = F(z) and F(x-1)+F(y-1)]=/=F(z-1) or vice versa So [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). Or F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). We have F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. So x^3+y^3=/=z^3. n>2. .Similar. We have a string inference G(z)*F(z)=G(x)*F(x)+G(y)*F(y) equivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) we see G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) give z=y. and G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 infer G(x)>0. give z=/=y. So G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) don't infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) don't infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) is not equiivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So have two cases [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) and [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) or vice versa. So [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. Or G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] We have x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] So x^n+y^n=/=z^n Happy&Peace. Trần Tấn Cường.

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