Please help 20 points
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<h2>Given :-</h2>
In □PQRS side PQ∥ side RS. If m∠P = 108degree
and m∠R = 53degree
<h2>To Find :-</h2>m∠Q and m∠S.
<h2>Solution :-</h2>According to angle sum property
P∠Q=180−∠P
∠Q=180−108
For angle S
∠S=180−∠R
∠S=180−53
<h2>2.</h2><h3>Given</h3>
- y=2x(2 -3x)
- y·y'' +x·y' -16 in simplest form
It is convenient to expand the expression for y to ease determination of derivatives.
... y = 4x -6x²
... y' = 4 -12x
... y'' = -12
Then the differential expression can be written as
... (4x -6x²)(-12) +x(4 -12x) -16
... = -48x +72x² +4x -12x² -16
... = 60x² -44x -16
<h2>3.</h2><h3>Given</h3>- y = 16/x +x³/3
- the turning points
- the extreme(s)
The derivative is
... y' = -16x^-2 + x^2
This is zero at the turning points, so
... -16/x^2 +x^2 = 0
... x^4 = 16 . . . . . . . . . multiply by x^2, add 16
... x^2 = ±√16 = ±4
We're only interested in the real values of x, so
... x = ±√4 = ±2 . . . . . . . x-values at the turning points
Then the turning points are
... y = 16/-2 +(-2)³/3 = -8 +-8/3 = -32/3 . . . . for x = -2
... y = 16/2 + 2³/3 = 8 +8/3 = 32/3 . . . . . . . for x = 2
The maximum is (-2, -10 2/3); the minimum is (2, 10 2/3).