Prove that: ${ \sqrt { n } }^{ { \sqrt { n } }^{ \sqrt { n } } }=\sqrt { { n }^{ \sqrt { { n }^{ \sqrt { n } } } } }$

Answers

Answer 1

Answer: Sorry for the quality :D

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Nells mortgage is 50,150 at 10 percent for 30 years. What is her monthly payment if she must pay 8.78 points per 1,000?A.) 440.32
B.) 439
C.) 422.95
D.) 385.89

Answers

I think it the answer is A.) 440.32

50,150 / 1,000 = 50.15
50.15 x 8.78 = 440.317 or 440.32

Which of the following functions has the greatest y-intercept?f(x)

x y
−3 −15
−2 −8
−1 −1
0 3
1 6
2 10
g(x) = −4 sin(5x) + 3

Answers

The y intercept occurs when x = 0

So for first function y-intercept = 3.

Second function:-

y intercept = -4 sin (5*0) + 3

= -4(0) + 3 = 3

y - intercepts are equal.

Solve the following expression when b = 12 10 + b/6 + b

Answers

Answer:

Step-by-step explanation:

Given that b = 12,

10 + 12/6 + 12

= 10 + 2 + 12

= 24

Answer:

Step-by-step explanation:

10+(12/6)+12 = 24 use brackets and divide first before adding

Which of the following statements are true? select all that apply

Answers

Answer:

its A :)

Step-by-step explanation:

The correct answer in the photo is A. B C D are wring

A number x is at least 40

Answers

Answer:

x ≥ 40

This reads: x is greater than or equal to 40. Meaning that x must be a number with a value of 40 or greater.

Step-by-step explanation:

Explain the derivation behind the derivative of sin(x) i.e. prove f'(sin(x)) = cos(x)How about cos(x) and tan(x)?

Answers

1.

$f'(\sin x) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\sin(x+h) - \sin(x))/(h) = \n \n = \lim_(h \to 0) (2 \sin( (x+h - x)/(2)) \cdot \cos( (x+h+x)/(2)) )/(h) = \lim_(h \to 0) (2 \sin( (h)/(2)) \cos( (2x+h)/(2) ) )/(h) = \n \n = \lim_(h \to 0) [ (\sin( (h)/(2)) )/( (h)/(2) ) \cdot \cos ((2x+h)/(2)) ] = \lim_(h \to 0) [1 \cdot \cos( (2x+h)/(2) ) ] =$

$= \cos( (2x)/(2)) = \boxed{\cos x}$

2.

$f'(\cos x) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\cos(x+h) - \cos(x))/(h) = \n \n = \lim_(h \to 0) (-2 \sin ( (x+h+x)/(2)) \cdot \sin ( (x+h-x)/(2)) )/(h) = \lim_(h \to 0) (-2 \sin ( (2x+h)/(2)) \cdot \sin ( (h)/(2)) )/(h) = \n \n = \lim_(h \to 0) (-2 \sin ( (2x+h)/(2)) )/(2) \cdot (sin( (h)/(2)) )/( (h)/(2) ) = \lim_(h \to 0) -\sin( (2x+h)/(2)) \cdot 1 =$

$= -\sin( (2x)/(2)) = \boxed{\sin x }$

3.

$f'(\tan) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\tan(x+h) - \tan(x))/(h) = \n \n = \lim_(h \to 0) ( (\sin(x+h-x))/(\cos(x+h) \cdot \cos(x)) )/(h) = \lim_(h \to 0) ( (\sin(h))/( (\cos(x+h-x) + \cos(x+h+x))/(2) ) )/(h) =$

$= \lim_(h \to 0) ( (\sin(h))/(\cos(h) + \cos(2x+h)) )/( (1)/(2)h ) = \lim_(h \to 0) (\sin(h))/( (1)/(2)h \cdot [\cos(h) + \cos(2x+h)] ) = \n \n = \lim_(h \to 0) (\sin(h))/(h) \cdot (1)/( (1)/(2) \cdot (\cos(h) + cos(2x+h) ) = 1 \cdot (1)/( (1)/(2) \cdot (1+ cos(2x) ) = (2)/(1 + 2 \cos^(2) - 1 ) = \n \n = (2)/(2 \cos^(2) x) = \boxed{ (1)/(\cos^(2)x) }$

4.

$f'(\cot) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\cot(x+h) - \cot(x))/(h) = \n \n = \lim_(h \to 0) ( (\sin(x - x - h))/(\sin (x+h) \cdot \sin (h)) )/(h) = \lim_(h \to 0) ( (\sin(-h) )/( (\cos(x+h-x) - \cos(x+h+x))/(2) ) )/(h) =$

$= \lim_(h \to 0) ( (-\sin(h))/(\cos(h) - \cos(2x+h)) )/( (1)/(2)h ) = \lim_(h \to 0) ( - \sin(h))/( (1)/(2)h \cdot [\cos(h) - \cos(2x+h)] ) = \n \n = \lim_(h \to 0) (- \sin (h))/(h) \cdot (1)/( (1)/(2) \cdot [\cos(h) - \cos(2x+h)] ) = -1 \cdot (2)/(1 - cos(2x)) = \n \n = - (2)/(1 -1 + 2 \sin^(2)x) = - (2)/(2 \sin^(2) x) = \boxed{- (1)/(\sin^(2) x) }$

I posted an image instead.

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