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AlgebraQuestion and Answers: Page 228

Question Number 119801 Answers: 0 Comments: 0

If f:R→R is a function such that f(0)=1 and f(x+f(y))= f(x)+y for all x, y∈R, then (A) 1 is a period of f (B) f(n)=1 for all integers n (C) f(n)=n for all integers n (D) f(−1)=0

$If f : R \to R is a function such that f (0) = 1 and f (x + f (y)) =$ $f (x) + y for all x, y \in R, then$ $(A) 1 is a period of f$ $(B) f (n) = 1 for all integers n$ $(C) f (n) = n for all integers n$ $(D) f (- 1) = 0$

Question Number 119797 Answers: 0 Comments: 0

Q1 Let M_2 be the set of square matrices of order 2 over the real number system and R={(A,B)∈M_2 ×M_2 ∣A=P^( T) BP for some non-singular P ∈M} Then R is (A) symmetric (B) transitive (C) reflexive on M_2 (D) not an equivalence relation on M_2 Q2 For any integer n, let I_n be the interval (n, n+1). Define R={(x, y)∈R∣both x, y ∈ I_n for some n∈Z} Then R is (A) reflexive on R (B) symmetric (C) transitive (D) an equivalence relation

$Q 1$ $Let M_{2} be the set of square matrices of order 2 over$ $the real number system and$ $R = {(A, B) \in M_{2} \times M_{2} ∣ A = P^{T} B P for some$ $non - singular P \in M}$ $Then R is$ $(A) symmetric$ $(B) transitive$ $(C) reflexive on M_{2}$ $(D) not an equivalence relation on M_{2}$ $Q 2$ $For any integer n, let I_{n} be the interval (n, n + 1) .$ $Define$ $R = {(x, y) \in R ∣ both x, y \in I_{n} for some n \in Z}$ $Then R is$ $(A) reflexive on R$ $(B) symmetric$ $(C) transitive$ $(D) an equivalence relation$

Question Number 119795 Answers: 4 Comments: 0

Solve in real numbers the system of equations { (((3x+y)(x+3y)(√(xy)) =14)),(((x+y)(x^2 +14xy+y^2 )= 36)) :}

$S o l v e i n r e a l n u m b e r s t h e s y s t e m o f$ $e q u a t i o n s {\begin{cases} (3 x + y) (x + 3 y) \sqrt{x y} = 14 \\ (x + y) (x^{2} + 14 x y + y^{2}) = 36 \end{cases}$

Question Number 119774 Answers: 1 Comments: 0

Question Number 119757 Answers: 0 Comments: 0

For any integer n, let I_n be the interval (n, n+1). Define R={(x, y)∈R∣both x, y ∈ I_n for some n∈Z} Then R is (A) reflexive on R (B) symmetric (C) transitive (D) an equivalence relation

$For any integer n, let I_{n} be the interval (n, n + 1) .$ $Define$ $R = {(x, y) \in R ∣ both x, y \in I_{n} for some n \in Z}$ $Then R is$ $(A) reflexive on R$ $(B) symmetric$ $(C) transitive$ $(D) an equivalence relation$

Question Number 119713 Answers: 1 Comments: 3

If 3x+(1/(2x))=6 find 8x^3 +(1/(27x^3 ))

$If 3 x + \frac{1}{2 x} = 6 find {8 x}^{3} + \frac{1}{{27 x}^{3}}$

Question Number 119635 Answers: 1 Comments: 0

If a function f:R→R satisfies the relation f(x+1)+f(x−1)=(√3)f(x) for all x∈R then a period of f is (A) 10 (B) 12 (C) 6 (D) 4

$If a function f : R \to R satisfies the relation$ $f (x + 1) + f (x - 1) = \sqrt{3} f (x) for all x \in R$ $then a period of f is$ $(A) 10 (B) 12 (C) 6 (D) 4$

Question Number 119634 Answers: 2 Comments: 0

Q1 If f:R→R is defined by f(x)=[x]+[x+(1/2)]+[x+(2/3)]−3x+5 where [x] is the integral part of x, then a period of f is (A) 1 (B) 2/3 (C) 1/2 (D) 1/3 Q2 Let a<c<b such that c−a=b−c. If f:R→R is a function satisfying the relation f(x+a)+f(x+b)=f(x+c) for all x∈R then a period of f is (A) (b−a) (B) 2(b−a) (C) 3(b−a) (D) 4(b−a)

$Q 1$ $If f : R \to R is defined by$ $f (x) = [x] + [x + \frac{1}{2}] + [x + \frac{2}{3}] - 3 x + 5$ $where [x] is the integral part of x, then a period of f is$ $(A) 1 (B) 2 / 3 (C) 1 / 2 (D) 1 / 3$ $Q 2$ $Let a < c < b such that c - a = b - c . If f : R \to R is a$ $function satisfying the relation$ $f (x + a) + f (x + b) = f (x + c) for all x \in R$ $then a period of f is$ $(A) (b - a) (B) 2 (b - a)$ $(C) 3 (b - a) (D) 4 (b - a)$

Question Number 119580 Answers: 3 Comments: 0

Given k ∈ N. 1) justify these relations: 3^(2k) +1≡2[8] and 3^(2k+1) +1≡4[8]. 2) Given (E): 2^n −3^m =1. n and m are unknowed. • Show that if m is even , (E) does not have solution. ■ Deduct from the first question 1) that the couple (2;1) is the only solution of (E).

$Given k \in N .$ $1) justify these relations :$ $3^{2 k} + 1 \equiv 2 [8] and 3^{2 k + 1} + 1 \equiv 4 [8] .$ $2) Given (E) : 2^{n} - 3^{m} = 1 . n and m are unknowed .$ $∙ Show that if m is even, (E) does not have$ $solution .$ $◼ Deduct from the first question 1) that the$ $couple (2; 1) is the only solution of (E) .$

Question Number 119576 Answers: 2 Comments: 0

Question Number 119540 Answers: 1 Comments: 0

Question Number 119538 Answers: 1 Comments: 1

Question Number 119490 Answers: 0 Comments: 0

Let a<c<b such that c−a=b−c. If f:R→R is a function satisfying the relation f(x+a)+f(x+b)=f(x+c) for all x∈R then a period of f is (A) (b−a) (B) 2(b−a) (C) 3(b−a) (D) 4(b−a)

$Let a < c < b such that c - a = b - c . If f : R \to R is a$ $function satisfying the relation$ $f (x + a) + f (x + b) = f (x + c) for all x \in R$ $then a period of f is$ $(A) (b - a) (B) 2 (b - a)$ $(C) 3 (b - a) (D) 4 (b - a)$

Question Number 119483 Answers: 1 Comments: 0

Let a>0 and f:R→R a function satisfying f(x+a)=1+[2−3f(x)+3f(x)^2 −f(x)^3 ]^(1/3) for all x∈R. Then a period of f(x) is ka where k is a positive integer whose value is (A)1 (B)2 (C)3 (D)4

$Let a > 0 and f : R \to R a function satisfying$ $f (x + a) = 1 + {[2 - 3 f (x) + 3 f {(x)}^{2} - f {(x)}^{3}]}^{1 / 3}$ $for all x \in R . Then a period of f (x) is k a where k is$ $a positive integer whose value is$ $(A) 1 (B) 2 (C) 3 (D) 4$

Question Number 119396 Answers: 3 Comments: 1

If the roots of the equation 24x^4 −52x^3 +18x^2 +13x−6=0 are α , −α , β and (1/β). Find the value of α and β.

$I f t h e r o o t s o f t h e e q u a t i o n$ $24 x^{4} - 52 x^{3} + 18 x^{2} + 13 x - 6 = 0 a r e$ $α, - α, β a n d \frac{1}{β} . F i n d t h e v a l u e o f$ $α a n d β .$

Question Number 119303 Answers: 3 Comments: 0

let x,y,z be positive real numbers such that x+y+z=1. Determine the minimum value of (1/x)+(4/y)+(9/z).

$l e t x, y, z b e p o s i t i v e r e a l n u m b e r s$ $s u c h t h a t x + y + z = 1 . D e t e r m i n e$ $t h e m i n i m u m v a l u e o f \frac{1}{x} + \frac{4}{y} + \frac{9}{z} .$

Question Number 119204 Answers: 0 Comments: 0

40−misolning yechimi: y=(x^3 /3)+2x^2 −5x+7 Kritik nuqtalarini topish uchun: 1. Funksiyadan hosila olamiz 2. Funksiya hosilasini nolga tenglab, tenglamani yechamiz. y′=x^2 +4x−5=0 ⇒ x_1 =1; x_2 =−5 x_1 +x_2 =1−5=−4 Javob: −4

$40 - misolning yechimi :$ $y = \frac{x^{3}}{3} + {2 x}^{2} - 5 x + 7$ $Kritik nuqtalarini topish uchun :$ $1 . Funksiyadan hosila olamiz$ $2 . Funksiya hosilasini nolga tenglab,$ $tenglamani yechamiz .$ $y^{'} = x^{2} + 4 x - 5 = 0 \Rightarrow x_{1} = 1; x_{2} = - 5$ $x_{1} + x_{2} = 1 - 5 = - 4 Javob : - 4$

Question Number 119160 Answers: 0 Comments: 5

Question Number 119159 Answers: 2 Comments: 0

We are in C. Given Z_(0 ) =1 ; Z_(n+1 ) =(1/2)Z_(n ) +(1/2)i n ∈ N. Show that ∀ n ∈ N^(∗ ) , ∣Z_n ∣<1.

$W e a r e i n C .$ $G i v e n Z_{0} = 1; Z_{n + 1} = \frac{1}{2} Z_{n} + \frac{1}{2} i$ $n \in N .$ $S h o w t h a t \forall n \in N^{*}, ∣ Z_{n} ∣< 1 .$

Question Number 119147 Answers: 0 Comments: 0

Question Number 119141 Answers: 1 Comments: 0

Question Number 119177 Answers: 1 Comments: 0

Question Number 119133 Answers: 0 Comments: 0

Informatica (11110000)_2 =0•2^0 +0•2^1 +0•2^2 +0•2^3 +1•2^4 +1•2^5 +1•2^6 +1•2^7 = =0•1+0•2+0•4+0•8+1•16+1•32+1•64+1•128= 0+0+0+0+16+32+64+128=(240)_(10) (11000101)_2 =1•2^0 +0•2^1 +1•2^2 +0•2^3 +0•2^4 +0•2^5 +1•2^6 +1•2^7 = = 1•1+0•2+1•4+0•8+0•16+0•32+1•64+1•128= (197)_(10) (01101001)_2 =

$Informatica$ ${(11110000)}_{2} = 0 ∙ 2^{0} + 0 ∙ 2^{1} + 0 ∙ 2^{2} + 0 ∙ 2^{3} + 1 ∙ 2^{4} + 1 ∙ 2^{5} + 1 ∙ 2^{6} + 1 ∙ 2^{7} =$ $= 0 ∙ 1 + 0 ∙ 2 + 0 ∙ 4 + 0 ∙ 8 + 1 ∙ 16 + 1 ∙ 32 + 1 ∙ 64 + 1 ∙ 128 =$ $0 + 0 + 0 + 0 + 16 + 32 + 64 + 128 = {(240)}_{10}$ ${(11000101)}_{2} = 1 ∙ 2^{0} + 0 ∙ 2^{1} + 1 ∙ 2^{2} + 0 ∙ 2^{3} + 0 ∙ 2^{4} + 0 ∙ 2^{5} + 1 ∙ 2^{6} + 1 ∙ 2^{7} =$ $= 1 ∙ 1 + 0 ∙ 2 + 1 ∙ 4 + 0 ∙ 8 + 0 ∙ 16 + 0 ∙ 32 + 1 ∙ 64 + 1 ∙ 128 = {(197)}_{10}$ ${(01101001)}_{2} =$

Question Number 119119 Answers: 2 Comments: 0

Question Number 119107 Answers: 1 Comments: 0

Question Number 119059 Answers: 1 Comments: 0

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