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Kevin C.'s Resources

Use the following identities:  cot x = 1/tanx. and tan 2x = (2tanx)/(1-tan2x)   The equation then becomes (2tanx)/(1-tan2x) - 1/(tanx) = 0   Add 1/(tanx) to each side:  (2tanx)/(1-tan2x) = 1/(tanx).   Multiply each side by tanx*(1-tan2x):  2tan2x...

If the problem is (x-2)2 = -8,  then |x-2| = 2i√2, and x-2 = 2i√2 and x-2 = -2i√2.   So the answers are x = 2+2i√2, and x = 2-2i√2.    

The Mean Value Thm states:  If f(x) is continuous over [a, b], and f'(x) is defined over (a, b), where a<<b, then there is a c such that f'(c) = (f(b) - f(a))/(b - a).  Try this and you should have your answer.   Therefore:  f'(c) = -14c,  and (f(5)-f(4))/(5-(-4))...

Hi, Carlos.   There are two ways to approach this problem.  We need to put this equation into a different form.   f(x) = a(x-h)2 + k, so the vertex will be the point (h, k).   One way is to complete the square:  f(x) = (x2 - 8x +   ...

Remember, if this is an inequality, if you divide by a negative number, the inequality changes.   Therefore, -9x + 9 >= 4 x - 8  becomes -13x >= -17.   Divide by -13, and the answer becomes:  x<= 17/13.

Answers solving equation (answer)

I don't see an equation.  However, if your equation is in the form:  1x2 + bx  = 0, in order to make sure the equation is a perfect square, divide b by 2, square it, and add to both sides.   Example:  x2 + 6x = 0,  b = 6, b/2 = 6/2 = 3,  so (b/2)2...

a)  The vertical asymptote is the value of x such that x-q = 0(i.e. when the f(x)-> ∞.      Therefore, x = q.        The horizontal asymptote is the value of f(x) when x-> ∞.       Therefore, y = 3 (the...

Again we have a perfect square.   Same process:  y2 - 2/3y + 1/9 = 0     (How do we know that this is a perfect square?   Take 1/2 of 2/3 = 1/3,  square  that and we get 1/9)   y2 - 2/3y + 1/9 = 0   (y -...

Note that the left side is a perfect square.  If we rewrite the left side as a square, all that is needed to do is take the square root of each side.   x2 + 10x + 25 = 81   (x+5)2 = 81   √(x + 5)2 = √81   |x + 5| = 9   Therefore, ...

Another approach:  6x2 - 19x + 10   Try to find 2 numbers whose product is 60  (the product of the coefficient of x2 and 10, the constant), and whose sum is 19.  Try 15 and 4.   Rewrite to get 6x2 - 15x - 4x + 10   Factor by grouping: ...

Using the equation,  y(t) = yoekt,  where y(t) is the value of y at time t, yo is the value of y at time  t = 0, and k is the rate at which y increases as a function of time.   Therefore, yo = 10,000 and k = .02.   a)  y(0) = 10,000e0.02*0 = 10,000(1)...

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