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I am not sure that the syntax is accurate. If the first equation should be

\(f(x)=({1\over2})^{2x}\)

then it is not linear. 

For x=0, f=1

For x=1, f=1/4

For x=2, f=1/16

If the second equation should be

\(f(x)=-{1\over{2}}*(x-2)\)

Then you can simplify to \(f(x)=-{x\over{2}}+1\)

which is linear. See that

For x=0, f=1

For x=1, f=1.5

For x=2, f=2

If asinus is correct then you can ignore this.

6/4 = 1.5

let f(x) = 6x^2 +x. Use the definition of the derivative as the limit of the difference quoetient to find f '(x).

\(\begin{array}{rcll} f(x) &=& 6x^2+x\\ f(x+h)&=&6(x+h)^2+(x+h)\\\\ \text{The difference quotient is:}\\ \dfrac{\Delta x}{\Delta y} &=& \dfrac{f(x+h)-f(x)}{h}\\ &=& \dfrac{6(x+h)^2+(x+h)-(6x^2+x)}{h}\\ &=& \dfrac{6(x+h)^2+ x +h-6x^2 - x }{h}\\ &=& \dfrac{6(x+h)^2 +h-6x^2 }{h}\\ &=& \dfrac{6(x^2+2xh+h^2) +h-6x^2 }{h}\\ &=& \dfrac{6x^2+12xh+6h^2 +h-6x^2 }{h}\\ &=& \dfrac{ 12xh+6h^2 +h }{h}\\ &=& \dfrac{ h( 12x+6h +1) }{h}\\ \dfrac{\Delta x}{\Delta y} &=& 12x+6h +1 \end{array}\)

\(\begin{array}{rcll} f'(x) &=& \dfrac{dy}{dx} \\ \dfrac{dy}{dx} &=& \lim \limits_{h\to 0} { \left( 12x+6h +1 \right) } \\ \dfrac{dy}{dx} &=& \left( 12x+6\cdot 0 +1 \right) \\ \dfrac{dy}{dx} &=& \left( 12x+1 \right) \\\\ \mathbf{f'(x)} &\mathbf{=}& \mathbf{12x+1} \end{array}\)

Find the derivative of the following via implicit differentiation:
d/dx(f(x)) = d/dx(x+6 x^2)
The derivative of f(x) is f'(x):
f'(x) = d/dx(x+6 x^2)
Differentiate the sum term by term and factor out constants:
f'(x) = d/dx(x)+6 d/dx(x^2)
The derivative of x is 1:
f'(x) = 6 (d/dx(x^2))+1
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x:
f'(x) = 1+6 2 x
Simplify the expression:
f'(x) = 1+12 x
Expand the left hand side:
Answer: |  f'(x) = 1+12x

It's fraction and means you have divided a Thing into 50 parts and have taken 20 parts out of 50.

Hi there, 

Nomenclature: the devising or choosing of names for things, especially in a science or other discipline.

Each 'Batch' of cookies calls for 3 cups of flour.....Liam only has 2 out of the 3 needed for a 'Batch'...or 2/3 the required amount. Lim should use only 2/3 of the required amount of sugar for a 'batch' .  Each 'batch' calls for 1 cup of sugar...Liam should use 2/3 's of 1 cup...or 2/3 cup sugar.

Hello Guest!

Which is linear

f(x) = (1/2)2x

or

f(x) = -1/2 (x-2) ?

f(x) = (1/2)2x

\(f(x) = \frac{1}{2}\times 2x \)

f(x) = x

f(x) is linear                      a straight {nl}  

f(x) = -1/2 (x-2)

\(f(x)= \frac{-1}{2(x-2)} \)

\(f(x) = \frac{ 1}{ 4- 2x}\)

f (x) is non-linear             a hyperbola {nl}  

Greeting asinus :- )

  !

3/8*4/5 = ?

3*4=12

8*5=40

=12/40

12/4=3

40/4=10

=3/10

=0.3 is your answer

60 - 42/14 = 57

Flour   5/1.5 = 3.33 'batches' of cookies

Sugar 4/1.25 = 3.2  ' batches'

Butter 1.75/.5 = 3.5 'batches'

The amount of sugar he has is the limiting ingredient      3.2  batches  x 15  cookies/batch  = 48  cookies

That is a good question.

I expect you already know the answer ;)

It's 60km/hour.      :))

Hi Cleopatrx,

It is a cool username you have there :)  Welcome to the forum:)

I am on my phone so this is not the best way to explain.

I looked for a fomula that linked number to cost. It is pretty silly because it keeps getting dearer.

Anyway if you square the number and multiply by 3 you get close to the cost.

If you then add 15 and subtract the production number it works out exact.

So

C= 3x^2 + 15 -x

C = 3x^2 - x +15

When x=5

C = 3 × 25 - 5 +15

C = 75 

If I understand you, you want the sum of the following A.S., A1=5, common difference=5, number of terms=50. If so, then the sum is=6,375.

Find the gradient of the radius that touches the tangent.

The gradient of the tangent is the negative reciprocal.

Use the point and the gradient to determine the equation.

\((\frac{1}{5}+\sqrt{\frac{19}{10i}})^2\\ = \frac{1}{25}+2\times \frac{1}{5}\times \sqrt{\frac{19}{10i}}+\frac{19}{10i}\\= \frac{1}{25}+\frac{2}{5}\times\frac{\sqrt{19}}{\sqrt{10}}\times \frac{\sqrt{2}}{1+i}+\frac{19}{10i}\)

Because sqrt i = (1+i)/ sqrt2

http://web2.0calc.com/questions/what-is-square-root-of-i

\(=\frac{1}{25}+\frac{2\sqrt{38}}{(5\sqrt{10})(1+i)}+\frac{19}{10i}\)

\(=\frac{1}{25}+\frac{2\sqrt{-380}}{50i(1+i)}+\frac{95+95i}{50i(1+i)}\)

\(=\frac{1}{25}+\frac{(2\sqrt{380}+95)i+95}{50i-50}\)

You can do the last half yourself.

That's the answer i am looking for.

Thx CPhill

\(2C^2=2(-2)^2=2\times 4 = 8\)

\(\log_2(1.05)\\\approx0.0704\)

2/5 + 1/2 = 4/10 + 5/10 = 9/10

6 1/4 - 3 2/3 = 6 3/12 - 3 8/12 = 2 7/12

(1 1/2)/(1/2)= 3/2 / 1/2 = 3

5/7 + 4/7 = 9/7 = 1 2/7

2 1/8 + 4 3/8 = 6 4/8 = 6 1/2

negative numbers.

p is false, q is true, so p ∧ q is false.

There's nothing you can do unless you are doing differentiations and integrations.