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LogicQuestion and Answers: Page 4

Question Number 93618 Answers: 0 Comments: 1

Question Number 92687 Answers: 0 Comments: 0

In a Gregorian calendar, a year finishing with 00 is a leap year if only it′s vintage is divisible by 400. Also, the 1^(st) January 1900 was a Monday. 1\ Show that a year with the vintage finishing with 00 cannot begin on a Sunday 2\ Show that for a person born between 1900−2071, his 28 anniversary will occur on the same day of birth.

$In a Gregorian calendar, a year finishing with$ $00 is a leap year if only {it}^{'} s vintage is divisible$ $by 400 . Also, the 1^{st} January 1900 was a$ $Monday .$ $1 ∖ Show that a year with the vintage finishing with$ $00 cannot begin on a Sunday$ $2 ∖ Show that for a person born between 1900 - 2071,$ $his 28 anniversary will occur on the same$ $day of birth .$

Question Number 92673 Answers: 1 Comments: 2

Show that if 3 prime numbers, all greater than 3, form an arithmetic progression then the common difference of the progression is divisible by 6.

$Show that if 3 prime numbers, all greater$ $than 3, form an arithmetic progression then the common$ $difference of the progression is divisible by 6 .$

Question Number 90801 Answers: 0 Comments: 3

what are the next two number of this series 4,7,9,2,5,6,3,8 ?

$w h a t a r e t h e n e x t t w o n u m b e r$ $o f t h i s s e r i e s 4, 7, 9, 2, 5, 6, 3, 8 ?$

Question Number 90233 Answers: 0 Comments: 5

3,4,12,39,103,x (a) 164 (b) 170 (c) 172 (d) 228

$3, 4, 12, 39, 103, x$ $(a) 164$ $(b) 170$ $(c) 172$ $(d) 228$

Question Number 90200 Answers: 0 Comments: 7

26≡R_1 [37] 1 ≡R_2 [3] 2≡R_3 [5] Find R_1 , R_2 and R_3

$26 \equiv R_{1} [37]$ $1 \equiv R_{2} [3]$ $2 \equiv R_{3} [5]$ $Find R_{1}, R_{2} and R_{3}$

Question Number 88943 Answers: 0 Comments: 2

Question Number 88940 Answers: 1 Comments: 0

Question Number 88690 Answers: 1 Comments: 0

Question Number 85555 Answers: 1 Comments: 0

2x^2 +5x+7=0

$2 x^{2} + 5 x + 7 = 0$

Question Number 79462 Answers: 1 Comments: 0

prove p⇒q and ∼q⇒∼p are logicaly equivalent with out truth table

$p r o v e p \Rightarrow q a n d \sim q \Rightarrow\sim p a r e l o g i c a l y$ $e q u i v a l e n t w i t h o u t t r u t h t a b l e$

Question Number 78671 Answers: 1 Comments: 0

prove p⇒q and negetion of q⇒negation of p

$p r o v e p \Rightarrow q a n d n e g e t i o n o f q \Rightarrow n e g a t i o n o f p$

Question Number 75255 Answers: 0 Comments: 0

Question Number 75254 Answers: 1 Comments: 0

Question Number 75257 Answers: 1 Comments: 0

Let A={2,4,6,7,8,9} B={1,3,5,6,10} and C={x:3x+6=0 or 2x+6=0}.Find a. A∪B. b. is(A∪B)∪C=A∪(B∪C)?

$L e t A = {2, 4, 6, 7, 8, 9}$ $B = {1, 3, 5, 6, 10} a n d$ $C = {x : 3 x + 6 = 0 o r 2 x + 6 = 0} . F i n d$ $a . A \cup B .$ $b . i s (A \cup B) \cup C = A \cup (B \cup C) ?$

Question Number 74246 Answers: 2 Comments: 0

Question Number 71849 Answers: 1 Comments: 0

Question Number 70742 Answers: 0 Comments: 1

Prove that f(x) = x − a cos (x) − b has at least one real root for ∀a,b ∈ R

$Prove that f (x) = x - a \cos (x) - b$ $has at least one real root for \forall a, b \in R$

Question Number 70168 Answers: 1 Comments: 3

Question Number 69116 Answers: 0 Comments: 0

Here are three propositions: − This sentence has exactly six words −There are two wrong propositions −The two previous sentences are correct Among that propositions ,how many are wrong? list them!

$H e r e a r e t h r e e p r o p o s i t i o n s :$ $- T h i s s e n t e n c e h a s e x a c t l y s i x w o r d s$ $- T h e r e a r e t w o w r o n g p r o p o s i t i o n s$ $- T h e t w o p r e v i o u s s e n t e n c e s a r e c o r r e c t$ $A m o n g t h a t p r o p o s i t i o n s, h o w m a n y a r e w r o n g ? l i s t t h e m!$

Question Number 68695 Answers: 1 Comments: 0

pour 1<k<n montrer que k(n+1−k)<(n+1/2)^2

$p o u r 1 < k < n m o n t r e r q u e$ $k (n + 1 - k) < {(n + 1 / 2)}^{2}$

Question Number 67651 Answers: 0 Comments: 0

1)Let consider S= Σ_(n=0) ^∞ n and T=Σ_(n=0) ^∞ (−1)^(n+1) n We know that ∀ x∈]−1;1] Σ_(n=0) ^∞ (−x)^n =(1/(1+x)) , then after derivating (1/((1+x)^2 ))=Σ_(n=1) ^∞ (−1)^(n+1) nx^(n−1) for x=1 ,we get T=(1/4) Now let ascertain something T=Σ_(n=0) ^∞ (−1)^(2n+2) (2n+1) +Σ_(n=0) ^∞ (−1)^(2n+1) (2n) =Σ_(n=0) ^∞ (2n+1) −2S (•) knowing that S=Σ_(n=0) ^∞ (2n)+Σ_(n=0) ^∞ (2n+1) So Σ_(n=0) ^∞ (2n+1)= S−2S When replacing that value in (•) we get T=(S−2S )−2S If i conclude that T=−3S and finally find S=−(T/3)=−(1/(12)) where will the mistake be? 2)Let consider K=1+((2020)/(2019))+(((2020)/(2019)))^2 +(((2020)/(2019)))^3 +......+... ((2020)/(2019))K=((2020)/(2019))+(((2020)/(2019)))^2 +(((2020)/(2019)))^3 +.... K−1=((2020)/(2019))+(((2020)/(2019)))^2 +(((2020)/(2019)))^3 +... So ((2020)/(2019))K=K−1 Then K=−2019 Where is the error? 3)Let consider n an integer We have 0=n−n =n+(−n)=n+(−n)^(2×(1/2)) =n+[(−n)^2 ]^(1/2) =n+ (√n^2 ) = n+n=2n So << all integer are null : 0 is the only integer>> Where is the error? 4) let consider n an integer different of zero and f(n)=nln(n) we have (df/dn)=ln(n)+1 (•) Likewise f(n)=ln(n^n ) and we know that n^n =n×n×n×......×n (n times) So f(n)=ln(n)+ln(n)+......+ln(n) (n times) Now we have (df/dn)=(1/n)+(1/n)+....+(1/n) (n times) So (df/( dn))=1 (••) Relation (•) and (••) give ln(n)+1=1 then ln(n)=0 ⇒ n=1 << The logarithm of all n≥1 is null : There is no integer big than 1 >> Where is the error? 5) let consider x=0,999999999....... we ascertain that 10x=9,999999999...... then 10x=9+0,999999999.... So 10x=9+x finally x=1 << 0.9999999999999999 ..... is and integer >> Is there any error? 6)let consider a=((26666666666666666)/(66666666666666665)) b=((999999999999999999999995)/(199999999999999999999999)) In the way to cancel , if i just remove one common figure to the numerator and to the denominator And i find a=(2/5) and b=(5/1) Will it be wrong? if no ,explain!

$1) L e t c o n s i d e r S = \underset{n = 0}{\sum^{\infty}} n a n d T = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n + 1} n$ $W e k n o w t h a t \forall x \in] - 1; 1]$ $\underset{n = 0}{\sum^{\infty}} {(- x)}^{n} = \frac{1}{1 + x},$ $t h e n a f t e r d e r i v a t i n g$ $\frac{1}{{(1 + x)}^{2}} = \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} {n x}^{n - 1}$ $f o r x = 1, w e g e t T = \frac{1}{4}$ $N o w l e t a s c e r t a i n s o m e t h i n g$ $T = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{2 n + 2} (2 n + 1) + \underset{n = 0}{\sum^{\infty}} {(- 1)}^{2 n + 1} (2 n)$ $= \underset{n = 0}{\sum^{\infty}} (2 n + 1) - 2 S (∙)$ $k n o w i n g t h a t S = \underset{n = 0}{\sum^{\infty}} (2 n) + \underset{n = 0}{\sum^{\infty}} (2 n + 1)$ $S o \underset{n = 0}{\sum^{\infty}} (2 n + 1) = S - 2 S$ $W h e n r e p l a c i n g t h a t v a l u e i n (∙)$ $w e g e t$ $T = (S - 2 S) - 2 S$ $I f i c o n c l u d e t h a t T = - 3 S$ $a n d f i n a l l y f i n d S = - \frac{T}{3} = - \frac{1}{12}$ $w h e r e w i l l t h e m i s t a k e b e ?$ $2) L e t c o n s i d e r K = 1 + \frac{2020}{2019} + {(\frac{2020}{2019})}^{2} + {(\frac{2020}{2019})}^{3} + . . . . . . + . . .$ $\frac{2020}{2019} K = \frac{2020}{2019} + {(\frac{2020}{2019})}^{2} + {(\frac{2020}{2019})}^{3} + . . . .$ $K - 1 = \frac{2020}{2019} + {(\frac{2020}{2019})}^{2} + {(\frac{2020}{2019})}^{3} + . . .$ $S o \frac{2020}{2019} K = K - 1$ $T h e n K = - 2019$ $W h e r e i s t h e e r r o r ?$ $3) L e t c o n s i d e r n a n i n t e g e r$ $W e h a v e$ $0 = n - n$ $= n + (- n) = n + {(- n)}^{2 \times \frac{1}{2}}$ $= n + {[{(- n)}^{2}]}^{\frac{1}{2}} = n + \sqrt{n^{2}} = n + n = 2 n$ $S o << a l l i n t e g e r a r e n u l l : 0 i s t h e o n l y i n t e g e r >>$ $W h e r e i s t h e e r r o r ?$ $4) l e t c o n s i d e r n a n i n t e g e r d i f f e r e n t o f z e r o a n d f (n) = n l n (n)$ $w e h a v e \frac{d f}{d n} = l n (n) + 1 (∙)$ $L i k e w i s e f (n) = l n (n^{n}) a n d w e k n o w t h a t$ $n^{n} = n \times n \times n \times . . . . . . \times n (n t i m e s)$ $S o f (n) = l n (n) + l n (n) + . . . . . . + l n (n) (n t i m e s)$ $N o w w e h a v e \frac{d f}{d n} = \frac{1}{n} + \frac{1}{n} + . . . . + \frac{1}{n} (n t i m e s)$ $S o \frac{d f}{d n} = 1 (∙ ∙)$ $R e l a t i o n (∙) a n d (∙ ∙) g i v e$ $l n (n) + 1 = 1 t h e n l n (n) = 0 \Rightarrow n = 1$ $<< T h e l o g a r i t h m o f a l l n ⩾ 1 i s n u l l : T h e r e i s n o i n t e g e r b i g t h a n 1 >>$ $W h e r e i s t h e e r r o r ?$ $5) l e t c o n s i d e r x = 0, 999999999 . . . . . . .$ $w e a s c e r t a i n t h a t$ $10 x = 9, 999999999 . . . . . .$ $t h e n 10 x = 9 + 0, 999999999 . . . .$ $S o 10 x = 9 + x$ $f i n a l l y x = 1$ $<< 0.9999999999999999 . . . . . i s a n d i n t e g e r >>$ $I s t h e r e a n y e r r o r ?$ $6) l e t c o n s i d e r a = \frac{26666666666666666}{66666666666666665}$ $b = \frac{999999999999999999999995}{199999999999999999999999}$ $I n t h e w a y t o c a n c e l, i f i j u s t r e m o v e o n e c o m m o n$ $f i g u r e t o t h e n u m e r a t o r a n d t o t h e d e n o m i n a t o r$ $A n d i f i n d a = \frac{2}{5} a n d b = \frac{5}{1}$ $W i l l i t b e w r o n g ? i f n o, e x p l a i n!$

Question Number 65427 Answers: 0 Comments: 1

Question Number 61934 Answers: 1 Comments: 2

Answer: 0^0 =?

$Answer : 0^{0} = ?$

Question Number 60484 Answers: 1 Comments: 2

If a + b + c = 4 then find a^3 + b^3 + c^(3 ) = ?

$I f a + b + c = 4$ $t h e n f i n d$ $a^{3} + b^{3} + c^{3} = ?$

Question Number 55057 Answers: 0 Comments: 0

Prove that Ln(z+1)=z−(z^2 /2)+(z^3 /3)−(z^4 /4)+... for ∣z∣< 1

$Prove that Ln (z + 1) = z - \frac{z^{2}}{2} + \frac{z^{3}}{3} - \frac{z^{4}}{4} + . . .$ $for ∣ z ∣< 1$

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