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Question Number 157739 Answers: 1 Comments: 1

two parallel paths 26m apart run east−west through the woods. one person east on one path at 3 kmph and another west on other path at 4 kmph. if they pass each other at time t =0 , i)how far apart are they 8 sec later? ii) how fast is the distance between them changing at that mlment? iii)find the distance L between them at 8 sec and (dL/dt)?

$t w o p a r a l l e l p a t h s 26 m a p a r t r u n$ $e a s t - w e s t t h r o u g h t h e w o o d s .$ $o n e p e r s o n e a s t o n o n e p a t h a t$ $3 k m p h a n d a n o t h e r w e s t o n o t h e r$ $p a t h a t 4 k m p h . i f t h e y p a s s e a c h$ $o t h e r a t t i m e t = 0,$ $i) h o w f a r a p a r t a r e t h e y$ $8 s e c l a t e r ?$ $i i) h o w f a s t i s t h e$ $d i s t a n c e b e t w e e n t h e m c h a n g i n g$ $a t t h a t m l m e n t ?$ $i i i) f i n d t h e d i s t a n c e L$ $b e t w e e n t h e m a t 8 s e c a n d \frac{d L}{d t} ?$

Question Number 157730 Answers: 0 Comments: 8

Good day Mr.W, I would have loved to send this privately to you but there′s nomeans of doing that. You assisted me in solving a problem and the answers are really correct; I was also understandinv the approach however some points got me lost. I′ll be most grateful if you could help me amplify the solution. I always appreciate you since I was here in 2014. Thank you now and always. here are some of the issues i have with the solution. I thought that in the composite system the axial load would be shared among the steel and concrete thus having a portion of the load in the steel and the rest in the concrete, yet you solved for the stress in concrete and used the general axial load and total area and yet your answer was very correct but i really dont understand why and how you had to use the general area and load for just concrete. I was also shocked about how the area of the steel became had α in it. I thought it would just be ((nπd^2 )/4). I would lie if I say I dont marvel at your calculation skills sir. please explain this method for me. It is really short and concise.The other method I know also gave me the answer you got but its a really long journey. Please sir, explain in detail to me. I′m most grateful. Thanks in advance. I′ll post the auestion and your previous workings.

$G o o d d a y M r . W, I w o u l d h a v e l o v e d$ $t o s e n d t h i s p r i v a t e l y t o y o u b u t$ ${t h e r e}^{'} s n o m e a n s o f d o i n g t h a t . Y o u$ $a s s i s t e d m e i n s o l v i n g a p r o b l e m$ $a n d t h e a n s w e r s a r e r e a l l y c o r r e c t; I$ $w a s a l s o u n d e r s t a n d i n v t h e a p p r o a c h$ $h o w e v e r s o m e p o i n t s g o t m e l o s t .$ $I^{'} l l b e m o s t g r a t e f u l i f y o u c o u l d$ $h e l p m e a m p l i f y t h e s o l u t i o n .$ $I a l w a y s a p p r e c i a t e y o u s i n c e I w a s$ $h e r e i n 2014 .$ $T h a n k y o u n o w a n d a l w a y s .$ $h e r e a r e s o m e o f t h e i s s u e s$ $i h a v e w i t h t h e s o l u t i o n .$ $I t h o u g h t t h a t i n t h e c o m p o s i t e$ $s y s t e m t h e a x i a l l o a d w o u l d b e$ $s h a r e d a m o n g t h e s t e e l a n d$ $c o n c r e t e t h u s h a v i n g a p o r t i o n$ $o f t h e l o a d i n t h e s t e e l a n d$ $t h e r e s t i n t h e c o n c r e t e, y e t$ $y o u s o l v e d f o r t h e s t r e s s i n$ $c o n c r e t e a n d u s e d t h e g e n e r a l a x i a l l o a d a n d$ $t o t a l a r e a a n d y e t y o u r a n s w e r$ $w a s v e r y c o r r e c t b u t i r e a l l y d o n t$ $u n d e r s t a n d w h y a n d h o w y o u h a d t o$ $u s e t h e g e n e r a l a r e a a n d l o a d f o r$ $j u s t c o n c r e t e .$ $I w a s a l s o s h o c k e d a b o u t h o w t h e$ $a r e a o f t h e s t e e l b e c a m e h a d α i n$ $i t . I t h o u g h t i t w o u l d j u s t b e \frac{n π d^{2}}{4} .$ $I w o u l d l i e i f I s a y I d o n t m a r v e l$ $a t y o u r c a l c u l a t i o n s k i l l s s i r .$ $p l e a s e e x p l a i n t h i s m e t h o d f o r m e .$ $I t i s r e a l l y s h o r t a n d c o n c i s e . T h e$ $o t h e r m e t h o d I k n o w a l s o g a v e m e$ $t h e a n s w e r y o u g o t b u t i t s a r e a l l y l o n g$ $j o u r n e y . P l e a s e s i r, e x p l a i n i n d e t a i l$ $t o m e . I^{'} m m o s t g r a t e f u l .$ $T h a n k s i n a d v a n c e .$ $I^{'} l l p o s t t h e a u e s t i o n a n d y o u r p r e v i o u s w o r k i n g s .$

Question Number 157725 Answers: 1 Comments: 0

form f(x,y,z) =((xy)′c)′((x′+c)(y′+z′))′ in standard SOP form and canonical SOP form

$form f (x, y, z) = {({(xy)}^{'} c)}^{'} {((x^{'} + c) (y^{'} + z^{'}))}^{'}$ $in standard SOP form and canonical SOP form$

Question Number 157724 Answers: 1 Comments: 2

8. A number can be expressed as a terminating decimal,if the denominator has factors : (a) 2,3 or 5 (b) 2 or 3 (c) 3 or 5 (d) 2 or 5 9. Given that : HCF of 2520 and 6600= 120, LCM of 2520 and 6600= 252×k, then the value of k is : (a) 165 (b) 1625 (c) 550 (d) 600 10. The decimal expansion of the rational number ((47)/(2^4 ×5^(3 ) )) will terminate after : (a) 3 places (b) 4 places (c) 5 places (d) 1 place 11. The perimeter of two similar triangles ABC and LMN are 60 cm and 48 cm respectively . If LM = 8cm,then lenght of AB is : (a) 10 cm (b) 8 cm (c) 5 cm (d) 6 cm 12. Ratio in which the line segment joining (1,−7) and (6,4) are divided by x-axis is given as: (a) 4 :7 (b) 2 : 5 (c) 7 : 4 (d) 5 : 2 13. 119^2 − 111^2 is : (a) Prime number (b) Composite number ( c) An odd composite number (d)An odd prime number 14. Side of square , whose diagonal is 16 cm is given by: (a) 6(√(2 )) cm (b) 4(√2) cm (c) 7(√(2 ))cm (d) 8(√(2 ))cm

$8 . A number can be expressed as a terminating decimal, if the denominator has factors :$ $(a) 2, 3 or 5$ $(b) 2 or 3$ $(c) 3 or 5$ $(d) 2 or 5$ $9 . Given that : HCF of 2520 and 6600 = 120, LCM of 2520 and 6600 = 252 \times k, then the value of k is :$ $(a) 165$ $(b) 1625$ $(c) 550$ $(d) 600$ $10 . The decimal expansion of the rational number \frac{47}{2^{4} \times 5^{3}} will terminate after :$ $(a) 3 places$ $(b) 4 places$ $(c) 5 places$ $(d) 1 place$ $11 . The perimeter of two similar triangles ABC and LMN are 60 cm and 48 cm respectively . If LM = 8 cm, then lenght of AB is :$ $(a) 10 cm$ $(b) 8 cm$ $(c) 5 cm$ $(d) 6 cm$ $12 . Ratio in which the line segment joining (1, - 7) and (6, 4) are divided by x - axis is given as :$ $(a) 4 : 7$ $(b) 2 : 5$ $(c) 7 : 4$ $(d) 5 : 2$ $13 . 119^{2} - 111^{2} is :$ $(a) Prime number$ $(b) Composite number$ $(c) An odd composite number$ $(d) An odd prime number$ $14 . Side of square, whose diagonal is 16 cm is given by :$ $(a) 6 \sqrt{2} cm$ $(b) 4 \sqrt{2} cm$ $(c) 7 \sqrt{2} cm$ $(d) 8 \sqrt{2} cm$

Question Number 157749 Answers: 0 Comments: 10

a^2 + b^2 + c^2 + d^2 = 4 a,b,c,d ∈ R max{a^3 + b^3 + c^3 + d^3 } = ?

$a^{2} + b^{2} + c^{2} + d^{2} = 4$ $a, b, c, d \in R$ $m a x {a^{3} + b^{3} + c^{3} + d^{3}} = ?$

Question Number 157750 Answers: 1 Comments: 0

∫_0 ^∞ ((1−cos 4x)/(xe^x )) dx=?

$\int_{0}^{\infty} \frac{1 - \cos 4 x}{{x e}^{x}} d x = ?$

Question Number 157793 Answers: 0 Comments: 0

Question Number 157712 Answers: 2 Comments: 2

Question Number 157706 Answers: 1 Comments: 0

Question Number 157701 Answers: 2 Comments: 0

a;b;c∈N (1/(a + (1/(b + (1/c))))) = ((16)/(37)) ⇒ a+b+c=?

$a; b; c \in N$ $\frac{1}{a + \frac{1}{b + \frac{1}{c}}} = \frac{16}{37} \Rightarrow a + b + c = ?$

Question Number 157695 Answers: 4 Comments: 0

Question Number 157694 Answers: 1 Comments: 0

x^3 =x+c ; 0<c≤(2/(3(√3))) find x, without trigonometric cubic formula.

$x^{3} = x + c; 0 < c ⩽ \frac{2}{3 \sqrt{3}}$ $f i n d x, w i t h o u t t r i g o n o m e t r i c$ $c u b i c f o r m u l a .$

Question Number 157688 Answers: 0 Comments: 0

lim_(x→0) ((1/(ln (x+(√(x^2 +1))))) −(1/(ln (x+1))) )=?

$\lim_{x \to 0} (\frac{1}{\ln (x + \sqrt{x^{2} + 1})} - \frac{1}{\ln (x + 1)}) = ?$

Question Number 157687 Answers: 2 Comments: 1

if A ; a ; b ∈ Z^+ and 6∙40!=A∙2^a ∙3^b find (a+b)_(max) = ?

$if A; a; b \in Z^{+} and 6 \cdot 40! = A \cdot 2^{a} \cdot 3^{b}$ $find {(a + b)}_{\max} = ?$

Question Number 157686 Answers: 3 Comments: 0

if f(x+1)-f(x)=3 and f(25)=72 find f(2) = ?

$if f (x + 1) - f (x) = 3 and f (25) = 72$ $find f (2) = ?$

Question Number 157665 Answers: 2 Comments: 0

Question Number 157664 Answers: 0 Comments: 3

{ ((ax+by=7)),((ax^2 +by^2 =49)),((ax^3 +by^3 =133)),((ax^4 +by^4 =406)) :} ⇒2014x+2014y−100a−100b−2014xy=?

${\begin{cases} a x + b y = 7 \\ {a x}^{2} + {b y}^{2} = 49 \\ {a x}^{3} + {b y}^{3} = 133 \\ {a x}^{4} + {b y}^{4} = 406 \end{cases}$ $\Rightarrow 2014 x + 2014 y - 100 a - 100 b - 2014 x y = ?$

Question Number 157663 Answers: 2 Comments: 0

Question Number 157660 Answers: 0 Comments: 0

Given x_1 = 1, x_2 , x_3 , …, is a real numbers sequence for n ≥ 1 with recurrence relation x_(n+1) − x_n = (1/(2x_n )) . [x] is expressed as the largest integer of x . [25x_(625) ] = ?

$G i v e n x_{1} = 1, x_{2}, x_{3}, \dots, i s a r e a l n u m b e r s s e q u e n c e f o r n ⩾ 1 w i t h$ $r e c u r r e n c e r e l a t i o n x_{n + 1} - x_{n} = \frac{1}{2 x_{n}} .$ $[x] i s e x p r e s s e d a s t h e l a r g e s t i n t e g e r o f x .$ $[25 x_{625}] = ?$