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It is -18.33... (repeating) deg C.

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85 F ahrenheit are how many Celsius?

85ºF = 29.44ºC


The temperature was 6f ahrenheit before it dropped 10 Fahrenheit what was the temperature after the drop?

The temperature after the drop was -4°F.


350 ahrenheit is what Celsius?

350ºF = 176.67ºC


What is 98.6 degrees ahrenheit in Celsius?

98.6 degrees Celsius is 37.1 degrees Fahrenheit.


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


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).


What is the third term of the sequence defined by the recursive rule f(1)2 f(n) f(n-1) plus 1?

To find the third term of the sequence defined by the recursive rule ( f(1) = 2 ) and ( f(n) = f(n-1) + 1 ), we first calculate ( f(2) ) using the recursive formula. Since ( f(1) = 2 ), we have ( f(2) = f(1) + 1 = 2 + 1 = 3 ). Next, we calculate ( f(3) ) as ( f(3) = f(2) + 1 = 3 + 1 = 4 ). Thus, the third term of the sequence is ( f(3) = 4 ).


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


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.