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iren2701 [21]
3 years ago
12

Please solve this question.

Mathematics
1 answer:
andre [41]3 years ago
8 0
  • <em>Answer:</em>

<em>x = - 2</em>

<em>y = 5</em>

  • <em>Step-by-step explanation:</em>

<em>Hi there ! </em>

<em>2x + 9 = y</em>

<em>y = - 4x - 3</em>

<em> </em>

<em>equality</em>

<em>2x + 9 = - 4x - 3</em>

<em>2x + 4x = - 9 - 3</em>

<em>6x = - 12</em>

<em>x = - 12 : 6</em>

<em>x = - 2</em>

<em>replace x = - 2</em>

<em>2(-2) + 9 = y</em>

<em>- 4 + 9 = y</em>

<em>y = 5</em>

<em>Good luck !</em>

<em />

<em />

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Answer:

A.

Step-by-step explanation:

so, the equation is

h(t) = -t² + 7t

so, we need to find the solutions for t (the time when the ball is exactly 10 ft in the air). there had to be 2 solutions, as the ball first goes up passing the 10 ft height, and then comes back down again, passing the 10 ft mark a second time. and between these 2 times the ball is higher (but not equal, so, we can only use < or > as inequality signs) than 10 ft.

10 = -t² + 7t

-t² + 7t - 10 = 0

the generation solution to a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = t

a = -1

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c = -10

t = (-7 ± sqrt(7² - 4×-1×-10))/(2×-1) =

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t1 = (-7 + 3)/-2 = -4/-2 = 2 seconds

t2 = (-7 - 3)/-2 = -10/-2 = 5 seconds

so, between 2 and 5 seconds airtime the ball is higher than 10 ft.

and remember : HIGHER THAN.

so, we cannot use any equality (like <= or >=).

t must be higher than 2 and lower than 5 :

2 < t < 5

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