MaxWong

The definition of derivative is \(\displaystyle\lim_{\Delta x \rightarrow 0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\)

Let f(x) = 4x^2 - 3x.

\(\quad f'(x)\\ =\displaystyle\lim_{\Delta x \rightarrow 0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\\ =\displaystyle\lim_{\Delta x \rightarrow 0}\dfrac{(4(x+\Delta x)^2 - 3(x + \Delta x))-(4x^2-3x)}{\Delta x}\\ =\displaystyle\lim_{\Delta x \rightarrow 0}\dfrac{4x^2+4\Delta x^2 + 8x\Delta x - 3x - 3\Delta x-4x^2+3x}{\Delta x}\\ =\displaystyle\lim_{\Delta x \rightarrow 0}\dfrac{4\Delta x^2 + 8x\Delta x - 3\Delta x}{\Delta x}\\ =\displaystyle\lim_{\Delta x \rightarrow 0}4\Delta x + 8x - 3\\ = 4(0) + 8x + 3\\ = 8x + 3\)

\(7\bowtie g\\ = 7 + \sqrt{g+\sqrt{g+\sqrt{g+\cdot\cdot\cdot}}}\)

That means the whole thing after the 7 must be equal 2.

The dots means extended to infinity.

Therefore let \(x=\sqrt{g+\sqrt{g+\sqrt{g+\cdot\cdot\cdot}}}\)(it equals 2 at the same time.)

\(\sqrt{g+x} = x\\\sqrt{g+2} = 2\\ g + 2 = 4\\ g = 2\)

Finished!!

I think you may know factorization of quadratics so I skipped the steps.

\(\quad n^2 + n - 20\\ = n^2 + 5n - 4n - 20\\ = n(n+5) -4(n+5)\\ =(n-4)(n+5)\)

The angle isn't the problem, the main thing is how far the object travels.

Total distance of the object travels 

= 100 + 50 + 50 + 170

= 370 ft

Speed of the object

\(=\dfrac{370ft}{5min}\\ = 74 ft/min\)

Example: you have y = 3x + 5

y = 3x + 5

y - 5 = 3x

x = y/3 - 5/3

Therefore the x-intercept is -5/3.

\(x + (1/5) x + (1/5)x + x = 180\\( 2 + (2/5))x =180\\ x = \dfrac{180}{\frac{12}{5}}=\dfrac{5}{12}\times 180 = 5 \times 15 = 75 \)

The expression is \(\dfrac{m}{2} + 7\)

2 is always 2.

I assume you mean 2x.

\(-\dfrac{4(3x+1)}{2(2x-4)}=\dfrac{1}{2}\\ \dfrac{4(3x+1)}{2(2x-4)}=-\dfrac{1}{2}\\ 24x + 8=8-4x\\ 28x = 0\\ x = 0\)

Therefore 2x = 0

You may mean \(2^x\) and \(2^x = 2^0 = 1\)

I have a more direct solution than you. 

a = a (vertically opposite angle) <-- This means the angle in the triangle is also 'a degrees'

a + b = 107 (exterior angle of triangle)

Finished!! :)

And n(n+1)/2 = 1 + 2 + ...... + n can be proved by mathematical induction.

Let P(n) be n(n+1)/2 = 1 + 2 + ....... + n

For n = 1,

LHS = 1(1+1)/2 = 1

RHS = 1 = LHS

P(1) is true.

Assume that P(k) is true,

For n = k + 1

 LHS 

= (k+1)(k+2)/2

 RHS 

= 1 + 2 + ...... + k + (k + 1) 

= (k + 1)(k)/2 + (k + 1)

= \((\dfrac{k+1}{2})(k + 2)\)

= (k+1)(k+2)/2

= LHS

Therefore when P(k) is true, P(k + 1) is true.

By the definition of mathematical induction, P(n) is true for all positive integers n.