Question

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( +quad r>) Parabola and points [ y=x ] [ begin{array}{l} text { Evolute } left(-4 t ; 3 t}+frac{1}{2}right) -10 leq t leq 10 \end{array} ] (4) ( \left(-4 t³3

Answer 1

The quadratic equation can be solved using the quadratic formula. The solutions for t are approximately 3.59 and -5.93.

Step-by-step explanation:

This expression is a quadratic equation of the form at² + bt + c = 0, where the constants are a = 4.90, b = -14.3, and c = -20.0. Its solutions are given by the quadratic formula:

t = (-b ± √(b² - 4ac)) / (2a)

Using the given values, we have:

t = (-(-14.3) ± √((-14.3)² - 4(4.90)(-20.0))) / (2(4.90))

Simplifying further, we get:

t = (14.3 ± √(204.49 + 392)) / 9.8

Therefore, the solutions for t are:

t = (14.3 + √596.49) / 9.8 ≈ 3.59

t = (14.3 - √596.49) / 9.8 ≈ -5.93