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OleMash [197]
2 years ago
10

The number T of teams that have participated in a robot-building competition for high-school students over a recent period of ti

me x(in years) can be modeled by T(x) = 17.155x^2 + 193.68x + 235.81, 0 ≤ x ≤ 6.
After how many years is the number of teams greater than 1000? Justify your answer.
Mathematics
1 answer:
Katarina [22]2 years ago
8 0

Solving <u>the quadratic equation</u>, it is found that the number of teams is <u>greater than 1000</u> after 3.1 years.

The <u>number of teams after x years </u>is given by:

T(x) = 17.155x^2 + 193.68x + 235.81

Initially, there are T(0) = 236 teams. If will 1000 be when T(x) = 1000, thus:

17.155x^2 + 193.68x + 235.81 = 1000

17.155x^2 + 193.68x - 764.19 = 0

Which has coefficients a = 17.155, b = 193.68, c = -764.19, then:

\Delta = (193.68)^2 - 4(17.155)(-764.19) = 89950.66

x_{1} = \frac{-193.68 + \sqrt{89950.66}}{2(17.155)} = 3.1

x_{2} = \frac{-193.68 - \sqrt{89950.66}}{2(17.155)} = -14.3  

We want the positive root, so the number of teams is greater than 1000 after 3.1 years.

A similar problem is given at brainly.com/question/10489198

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Answer:

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8. x = 2 or x = 6

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Step-by-step explanation:

7. Subtract 320 from both sides: 4(x + 2)^2 = -320

Divide by 4: (x + 2)^2 = -80

Square root both sides: x + 2 = +/- \sqrt{-80}. We need to add the imaginary i to this: +/- \sqrt{-80} = +/- i\sqrt{80} = +/- 4i\sqrt{5}

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9. Add 5 to both sides: -2(x + 2)^2 = 13

Divide by -2: (x + 2)^2 = -13/2

Square root both sides: x + 2 = +/- \sqrt{-13/2}. We again need i: +/- \sqrt{-13/2} = +/- i\sqrt{13/2} = +/- \frac{\sqrt{26} }{2} i

Subtract 2 from both sides: x = -2 +/- \frac{\sqrt{26} }{2} i

10. Multiply by 5 on both sides: (t + 3)^2 = 35

Square root both sides: t + 3 = +/- \sqrt{35}

Subtract 3: t = -3 +/- \sqrt{35}

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