9514 1404 393
Answer:
f'(x) = (-6x² -14x -23)/(x² +5x +2)²
f''(x) = (12x³ +42x² +138x +202)/(x² +5x +2)³
Step-by-step explanation:
The applicable derivative formula is ...
d(u/v) = (v·du -u·dv)/v²
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f'(x) = ((-x² -5x -2)(4x +4) -(2x² +4x -3)(-2x -5))/(-x² -5x -2)²
f'(x) = (-4x³ -24x²-28x -8 +4x³ +18x² +14x -15)/(x² +5x +2)²
f'(x) = (-6x² -14x -23)/(x² +5x +2)²
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Similarly, the second derivative is the derivative of f'(x).
f''(x) = ((x² +5x +2)²(-12x -14) -(-6x² -14x -23)(2(x² +5x +2)(2x +5)))/(x² +5x +2)⁴
f''(x) = ((x² +5x +2)(-12x -14) +2(6x² +14x +23)(2x +5))/(x² +5x +2)³
f''(x) = (12x³ +42x² +138x +202)/(x² +5x +2)³
Answer:
About 17.7 meters.
Step-by-step explanation:
This can be solved by imagining the triangle formed by the building and its shadow. The hypotenuse of the triangle, the distance from the tip of the building to the tip of the shadow, is 34 meters, and one of the legs is 29 meters. Therefore, we can use the Pythagorean theorem to find that the third side is . Hope this helps!
For #8, it's approximately 2.5
Answer:
5th year= 9800
Step-by-step explanation:
1st year=15800
2nd=14300
3rd=12800
4th=11300
5th=9800
