Alan

Sorry Melody, I didn't realise you were working on this.

This question seems to have appeared twice!  I answered the other one at https://web2.0calc.com/questions/help-please_18822#r1.

A partial fractions problem:

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"Find all values of  t such that \(\lfloor t\rfloor = 2t + 3\) . If you find more than one value, then list the values you find in increasing order, separated by commas."

This graph should help.  HOWEVER, be careful to check all the values of t that seem to be suggested by the graph!!

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As follows:

  \(\text{Start as follows: }\frac{2x-y}{x+y}=\frac{2}{3}\\ \text{ Then multiply both sides by }3(x+y)\\ 3(2x-y)=2(x+y)\text{ or }6x-3y=2x+2y\\\text{Collect like terms together: }4x=5y\\\text{Divide both sides by }4y\\\frac{x}{y}=\frac{5}{4}\)

As follows:

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Use Pythagoras:  (a+1)² + (6a)² = (6a+1)²

a² + 2a + 1 + (6a)² = (6a)² + 12a + 1

a² -10a =0 

a = 0 or a = 10.

Since we can’t have a triangle with length 0 (6a), we must have a = 10.

As x→0 sin(ax)→ax, so \(\lim_{x\rightarrow 0}\frac{3sin{4x}}{sin{3x}}\rightarrow\frac{3*4x}{3x}\rightarrow4\)

"Find the equation of the line tangent to the given curve at the indicated point.

y = x^2 + 8x at (2, 20)

--> I got y-20 = 10(x-2)

y = x^3 – 4x at (3, 15)"

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Alternatively, you could do it algebraically.  Equate the two expressions for y:

x²+k = 1-kx

x² + kx + k-1 = 0.  (1)

The discriminant  of this quadratic is k²-4*1*(k-1) or k²-4k+4.  This is zero when k = 2.

Since there is only a single solution to equation (1) when the discriminant is zero, k = 2 doesn’t allow the original two equations to intersect in more than one point.

The graph above might help (even though it is a little busy!).