All Answers 358150 Answers

Hi Yura_chan,

I've done the algebra below but you reall want to graph this and try to understand what is happening :/

f(X) = 2x^2 + 4 at x=1 Find the derivative, both a and b should have the same answer.

a.) Use equation lim as x -> c f(x)-f(c)/x-c 

\(\displaystyle \lim_{x\rightarrow1} \;\;\frac{f(x)-f(1)}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2x^2+4-6}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2x^2-2}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2(x^2-1)}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2(x-1)(x+1)}{x-1}\\ =\displaystyle \lim_{x\rightarrow1} \;\;\frac{2(x+1)}{1}\\ =2*(1+1)\\ =4\)

b.) Use equation lim as h ->0 f(c+h)-f(c)/h

\(\displaystyle \lim_{h\rightarrow0} \;\;\frac{f(c+h)-f(c)}{c+h-c}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{f(1+h)-f(1)}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2h^2+4h+6-6}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2h^2+4h}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2h(h+2)}{h}\\ =\displaystyle \lim_{h\rightarrow0} \;\;\frac{2(h+2)}{1}\\ =2*(0+2)\\ =4\)

this means that the gradient of the tangent to the curve f(x)=2x^2+4        is     4   when x=1

Order of operations - start with multiplication and then do the addition.

So 4*6 = 24

24+3=27

Final answer is 27

with? 

cute hair in your pic wish I knew how to do that, haha I would propabily end up looking like a poodle (as usual) typical me. Oh anyways had a so/so day, thanks!! (NOT SARCASM) Whenever I type it sounds like im being sarcastic in my head!! SORRY!!  I am so weird haha

(x+4)^2 + (y-3)^2 = 25: y = x + 2        sub the second equation into the first for y

(x + 4)^2  + ( x + 2 - 3)^2  = 25

(x + 4)^2 + (x - 1)^2  = 25     expand the left side

x^2 + 8x + 16 + x^2 - 2x + 1   = 25     simplify

2x^2 + 6x + 17 = 25      subtract 25 from both sides

2x^2 + 6x - 8  = 0      divide through by 2

x^2 + 3x - 4  =  0     factor

(x - 1) ( x + 4)  = 0       and setiing each factor to 0, we have that  x  = 1  and x = -4

So when x = 1, y= 1 + 2 = 3     and when x = -4,  y= -4 + 2  = -2

So....the intersection points are  (1, 3)  and  ( -4, -2)

We   need to find the radius......this is given by

sqrt [ (-2 - - 3)^2  + ( -3 - 3)^2  ]   =  sqrt [ 1^2 + (-6)^2]  =  sqrt(37)

So.........the equation is :

(x - x coordinate of the center)^2 + (y - y coordinate of the center)^2  =  radius^2

(x - - 3)^2  + ( y - - 3)^2   = 37

(x + 3)^2  + ( y + 3)^2   = 37

x^2 + y^2 = 25: y = x     

Put the second equation into the first for y   and we have

x^2 + x^2  = 25

2x^2  = 25    divide both sides by 2

x^2  = 25/2          take the square root of both sides

x^2  = [ +/-] 5/sqrt( 2)

And since x = y   the intersection points are :

(5/sqrt(2), 5/sqrt(2) )  and   (-5/sqrt(2), -5/sqrt(2) )

Remember, smoky......an inscribed angle = 1/2 of its intercepted arc...

10.    5x + 4 =   (1/2)(208)

         5x + 4 =  104      subtract 4 from both sides

         5x  = 100       divide by 5 on each side

          x  = 20

2.   Notice that if we connected A and B, we would have an isosceles triangle.....and each base angle would equal    [ 180  - 48] / 2   =  132/2  =  66°  = angle BAC = angle ABC

So   angle BAC  =  [1/2] x°

66 = (1/2)x     multiply both sides by 2

132°  = x

7.   The  90° angle in the right triangle intercepts an arc of twice its measure.....therefore,  x = 180°

And since the other two sides of the right triangle are equal, the angles opposite those sides are equal, too.....so  each angle = 45°  and each intercepts an arc twice its measure.....therefore, y = 90°

Thank you! I understand now :)

Nice!!!      And i thought that i was the only one 

You do not find the derivative of 3x=0

You find the derivative of   y=3x

\(y=3x^1\\ \frac{dy}{dx}=1*3x^{1-1}\\ \frac{dy}{dx}=3x^{0}\\ \frac{dy}{dx}=3 \)

Also remember that the derivative of the function is the gradient of the tangent at any given point.

y=3x is a line.  the gradient is ALWAYS 3

The derivative of e^x is e^x   so the derivative of  3e^x   IS    3e^x

Same for me, man!

the varieable is x and y

Thanks, gliu1  ..   however....I think we had about 20+ on here earlier today.....that's a LOT  !!!

Thanks. Hard to get comments like that. XD

The answer is 2.

C = (5/9)(F-32)      multiply both sides by 9/5

(9/5)C =  F - 32     add32 to both sides

(9/5)C + 32   = F

Your welcome and thanks Hayley :) !!!

Thanks! (/^▽^)/

Puppies?! 

Yay! (ﾉ◕ヮ◕)ﾉ*:･ﾟ✧ ✧ﾟ･: *ヽ(◕ヮ◕ヽ)

You too :)

I try to be nice sometimes... :)

Thanks, Hayley