what does mean that we say C^° =F^° in −40?


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Question Number 204230 by Davidtim last updated on 09/Feb/24

$w h a t d o e s m e a n t h a t w e s a y C^{°} = F^{°}$ $i n - 40 ?$
Commented by mr W last updated on 10/Feb/24

$g e n e r a l l y t h e t e m p e r a t u r e v a l u e i n$ $u n i t ° C (C e l s i u s) i s n o t t h e s a m e a s$ $i t s v a l u e i n u n i t ° F (F a h r e n h e i t),$ $f o r e x a m p l e 10 ° C = 50 ° F,$ $- 20 ° C = - 4 ° F . b u t - 40 ° C = - 40 ° F .$
Commented by Davidtim last updated on 10/Feb/24

$t h a n k s$
Commented by Davidtim last updated on 10/Feb/24

$w o u l d y o u m i n d f i n d t h e \frac{d}{d x} (x!) ?$
Commented by mr W last updated on 10/Feb/24

$x! = Γ (x + 1)$ $\frac{d (x!)}{d x} = x! ψ^{(0)} (x + 1)$
Commented by Davidtim last updated on 12/Feb/24

$w o u l d y o u m i n d t o s o l v e w i t h o u t g a m m a$ $o r b i t t a f u n c t i o n ?$
Commented by mr W last updated on 12/Feb/24

$i c a n n o t . i f y o u c a n, p l e a s e s h a r e .$
Commented by Davidtim last updated on 12/Feb/24

$y = x!$ $l n y = l n (x!)$ $l n y = l n [x \cdot (x - 1) \cdot (x - 2) \cdot (x - 3) \cdot \cdot \cdot 3 \cdot 2 \cdot 1]$ $l n y = l n x + l n (x - 1) + l n (x - 2) + l n (x - 3) + \cdot \cdot \cdot + l n 1$ $\frac{y^{^{'}}}{y} = \frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2} + \cdot \cdot \cdot + 0$ $y^{^{'}} = y (\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2} + \cdot \cdot)$ $y^{^{'}} = x! (\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2} + \cdot \cdot \cdot)$
Commented by Davidtim last updated on 12/Feb/24

$y = x!$ $l n y = l n (x!)$ $l n y = l n [x \cdot (x - 1) \cdot (x - 2) \cdot (x - 3) \cdot \cdot \cdot 3 \cdot 2 \cdot 1]$ $l n y = l n x + l n (x - 1) + l n (x - 2) + l n (x - 3) + \cdot \cdot \cdot + l n 1$ $\frac{y^{^{'}}}{y} = \frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2} + \cdot \cdot \cdot + 0$ $y^{^{'}} = y (\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2} + \cdot \cdot)$ $y^{^{'}} = x! (\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2} + \cdot \cdot \cdot)$
Commented by mr W last updated on 12/Feb/24

$w h e n y o u t a k e x! = x (x - 1) (x - 2) . . . 3 \times 2 \times 1$ $y o u a s s u m e x \in N, t h e n x! i s n o t$ $a c o n t i n u o u s f u n c t i o n a n d t h u s n o t$ $d i f f e r e n t i a b l e .$
Commented by Davidtim last updated on 13/Feb/24

$h o w c a n s o l v e w i t h o u t g a m m a f u n c t i o n ?$
	Answered by Satyam1234 last updated on 09/Feb/24

	$F = C \times \frac{9}{5} + 32$ $L e t C = F = x$ $x = \frac{9 x}{5} + 32$ $5 x = 9 x + 160$ $- 4 x = 160$ $x = - 40$
	Commented by Davidtim last updated on 09/Feb/24

	$I {d o n}^{'} t u n d e r s t a n t t h e m a i n m e a n o t$ $t h i s .$ $d e s c r i b e t h e m a i n m e a n o f t h e - 40 .$
	Commented by JDamian last updated on 10/Feb/24

	$the main purpose is to make you to solve$ $a first degree equation$
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