Question and Answers Forum

AllQuestion and Answers: Page 371

Question Number 176962 Answers: 0 Comments: 2

Recommended books for trigonometry from zero? I know most Euclidean geometry, and basic algebra. I feel a bit lost

$R e c o m m e n d e d b o o k s f o r t r i g o n o m e t r y f r o m$ $z e r o ? I k n o w m o s t E u c l i d e a n g e o m e t r y,$ $a n d b a s i c a l g e b r a . I f e e l a b i t l o s t$

Question Number 176950 Answers: 1 Comments: 0

Question Number 176948 Answers: 1 Comments: 2

Question Number 176946 Answers: 1 Comments: 0

Question Number 176942 Answers: 1 Comments: 0

Question Number 176936 Answers: 1 Comments: 0

Question Number 176933 Answers: 1 Comments: 0

Question Number 182216 Answers: 3 Comments: 2

Find a, b, c ∈ N ; 2^( a) + 4^( b) + 8^( c) = 328

$F i n d a, b, c \in N; 2^{a} + 4^{b} + 8^{c} = 328$

Question Number 176928 Answers: 1 Comments: 0

What is the smallest three digit number divisible by (2/3), (7/9), (4/(13)) without a remainder ?

$What is the smallest three digit number$ $divisible by \frac{2}{3}, \frac{7}{9}, \frac{4}{13} without a remainder ?$

Question Number 176925 Answers: 1 Comments: 0

Question Number 176917 Answers: 0 Comments: 0

Question Number 176915 Answers: 1 Comments: 0

Question Number 176913 Answers: 1 Comments: 1

Determiner la hauteur DE(r+x) en fonction de r r=OOC=BH BF=20 pour que distance(AB+BC+CD+DE+EF soit sgale AC+arcCDF

$D e t e r m i n e r l a h a u t e u r D E (r + x) e n f o n c t i o n d e r$ $r = OOC = BH BF = 20$ $pour que distance (AB + BC + CD + DE + EF soit$ $sgale AC + arcCDF$

Question Number 176898 Answers: 1 Comments: 0

Question Number 176746 Answers: 5 Comments: 0

a_(n+2) −5a_(n+1) +6a_n =3n+5^n a_1 =1, a_2 =0 find a_n

$a_{n + 2} - 5 a_{n + 1} + 6 a_{n} = 3 n + 5^{n}$ $a_{1} = 1, a_{2} = 0$ $f i n d a_{n}$

Question Number 176741 Answers: 1 Comments: 1

In the following figure, if ((S_1 +S_2 )/( (√S_3 ))) = (6/( (√π))) find the area of the circular crown

$I n t h e f o l l o w i n g f i g u r e, i f \frac{S_{1} + S_{2}}{\sqrt{S_{3}}} = \frac{6}{\sqrt{π}}$ $f i n d t h e a r e a o f t h e c i r c u l a r c r o w n$

Question Number 176727 Answers: 0 Comments: 0

In △ABC the following relationship holds: 6r Σ_(cyc) (r_a /(s + n_a )) + Σ_(cyc) n_a ≥ 3s

$In △ ABC the following relationship$ $holds :$ $6 r \sum_{cyc} \frac{r_{a}}{s + n_{a}} + \sum_{cyc} n_{a} ⩾ 3 s$

Question Number 176723 Answers: 0 Comments: 0

donner la forme trigonometrique(i−1)^5 /(i−4)^4

$d o n n e r l a f o r m e t r i g o n o m e t r i q u e {(i - 1)}^{5} / {(i - 4)}^{4}$

Question Number 176721 Answers: 2 Comments: 0

Question Number 176722 Answers: 0 Comments: 1

(((√(11))+1)/(2(√(11))+11)) = ((((√(11))+1)(2(√(11))−11))/((2(√(11))+11)(2(√(11))−11))) = ((2(√(11))−11+2×11+11(√(11)))/(2×11−121)) = ((13(√(11))+11)/(−99)) Ans

$\frac{\sqrt{11} + 1}{2 \sqrt{11} + 11} = \frac{(\sqrt{11} + 1) (2 \sqrt{11} - 11)}{(2 \sqrt{11} + 11) (2 \sqrt{11} - 11)} =$ $\frac{2 \sqrt{11} - 11 + 2 \times 11 + 11 \sqrt{11}}{2 \times 11 - 121} = \frac{13 \sqrt{11} + 11}{- 99}$ $A n s$

Question Number 176719 Answers: 0 Comments: 0

Question Number 176718 Answers: 1 Comments: 1

x^3 +y^3 =z^2 x^3 +z^3 =y^2 y^3 +z^3 =x^2 obviously x=y=z=0∨(1/2) trying to totally solve it let y=px∧z=qx (p^3 +1)x^3 =q^2 x^2 (q^3 +1)x^3 =p^2 x^2 (p^3 +q^3 )x^3 =x^2 ⇒ x=0 ⇒ y=0∧z=0 ★ x=(q^2 /(p^3 +1)) x=(p^2 /(q^3 +1)) x=(1/(p^3 +q^3 )) ⇒ (p^2 /(q^3 +1))=(q^2 /(p^3 +1)) (p^2 /(q^3 +1))=(1/(p^3 +q^3 )) ========== p^5 +p^2 −q^5 −q^2 =0 p^5 +p^2 q^3 −q^3 −1=0 subtracting both p^2 (q^3 −1)+q^5 −q^3 +q^2 −1=0 (q−1)(p^2 (q^2 +q+1)+q^4 +q^3 +q+1)=0 ⇒ q=1 ⇒ p^5 +p^2 −2=0 (p−1)(p^4 +p^3 +2p+2)=0 ⇒ p=1 ⇒ x=y=z=(1/2) ★ p^4 +p^3 +2p+2=0 no useable exact solutions p≈−.975564±.528237i ⇒ x≈.33635∓.515329i ∧ y≈−.0559113±.680406i ∧ z=x ★ p≈.475564±1.18273i ⇒ x≈−.586346±.562464i ∧ y≈−.944089∓.426001i ∧ z=x ★ p^2 (q^2 +q+1)+q^4 +q^3 +q+1=0 p^2 =−(((q+1)^2 (q^2 −q+1))/(q^2 +q+1)) we can be sure that (q≠−1⇒p=0) ⇒ p^2 <0 ⇒ p=±(√((q^2 −q+1)/(q^2 +q+1)))(q+1)i ...I′ll continue later

$x^{3} + y^{3} = z^{2}$ $x^{3} + z^{3} = y^{2}$ $y^{3} + z^{3} = x^{2}$ $obviously x = y = z = 0 \lor \frac{1}{2}$ $trying to totally solve it$ $let y = p x \land z = q x$ $(p^{3} + 1) x^{3} = q^{2} x^{2}$ $(q^{3} + 1) x^{3} = p^{2} x^{2}$ $(p^{3} + q^{3}) x^{3} = x^{2}$ $\Rightarrow x = 0 \Rightarrow y = 0 \land z = 0 ★$ $x = \frac{q^{2}}{p^{3} + 1}$ $x = \frac{p^{2}}{q^{3} + 1}$ $x = \frac{1}{p^{3} + q^{3}}$ $\Rightarrow$ $\frac{p^{2}}{q^{3} + 1} = \frac{q^{2}}{p^{3} + 1}$ $\frac{p^{2}}{q^{3} + 1} = \frac{1}{p^{3} + q^{3}}$ $==========$ $p^{5} + p^{2} - q^{5} - q^{2} = 0$ $p^{5} + p^{2} q^{3} - q^{3} - 1 = 0$ $subtracting both$ $p^{2} (q^{3} - 1) + q^{5} - q^{3} + q^{2} - 1 = 0$ $(q - 1) (p^{2} (q^{2} + q + 1) + q^{4} + q^{3} + q + 1) = 0$ $\Rightarrow q = 1 \Rightarrow p^{5} + p^{2} - 2 = 0$ $(p - 1) (p^{4} + p^{3} + 2 p + 2) = 0$ $\Rightarrow p = 1 \Rightarrow x = y = z = \frac{1}{2} ★$ $p^{4} + p^{3} + 2 p + 2 = 0$ $no useable exact solutions$ $p \approx - . 975564 \pm . 528237 i$ $\Rightarrow x \approx . 33635 \mp . 515329 i$ $\land y \approx - . 0559113 \pm . 680406 i$ $\land z = x ★$ $p \approx . 475564 \pm 1 . 18273 i$ $\Rightarrow x \approx - . 586346 \pm . 562464 i$ $\land y \approx - . 944089 \mp . 426001 i$ $\land z = x ★$ $p^{2} (q^{2} + q + 1) + q^{4} + q^{3} + q + 1 = 0$ $p^{2} = - \frac{{(q + 1)}^{2} (q^{2} - q + 1)}{q^{2} + q + 1}$ $we can be sure that (q \neq - 1 \Rightarrow p = 0)$ $\Rightarrow p^{2} < 0 \Rightarrow p = \pm \sqrt{\frac{q^{2} - q + 1}{q^{2} + q + 1}} (q + 1) i$ $. . . I^{'} ll continue later$

Question Number 176716 Answers: 1 Comments: 0

Question Number 176710 Answers: 3 Comments: 0

Question Number 176708 Answers: 1 Comments: 0

a_(n+2) −3a_(n+1) +2a_n =n+3^n please find a_n

$a_{n + 2} - 3 a_{n + 1} + 2 a_{n} = n + 3^{n}$ $p l e a s e f i n d a_{n}$

Question Number 176692 Answers: 1 Comments: 1

Pg 366 Pg 367 Pg 368 Pg 369 Pg 370 Pg 371 Pg 372 Pg 373 Pg 374 Pg 375

Contact: info@tinkutara.com