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Ana dives into a pool off of a springboard high dive. Her height (in meters above the water), \( x \) seconds after diving, is modeled by

[tex]\[ h(x) = -5(x + 1)(x - 3) \][/tex]

How many seconds after diving will Ana hit the water?

[tex]\[ \square \][/tex] seconds



Answer :

To determine how many seconds after diving Ana will hit the water, we need to find the time \( x \) when her height \( h(x) \) above the water is zero.

Given the equation modeling her height:

[tex]\[ h(x) = -5(x + 1)(x - 3) \][/tex]

we set \( h(x) \) to 0 and solve for \( x \):

[tex]\[ -5(x + 1)(x - 3) = 0 \][/tex]

Since the product of two factors is zero, at least one of the factors must be zero. Therefore, we set each factor to zero and solve for \( x \):

1. \( x + 1 = 0 \)

[tex]\[ x = -1 \][/tex]

2. \( x - 3 = 0 \)

[tex]\[ x = 3 \][/tex]

These calculations give us two potential solutions: \( x = -1 \) and \( x = 3 \).

However, in the context of this problem, \( x \) represents the time in seconds after Ana dives. Time cannot be negative, so we discard \( x = -1 \).

Thus, the valid solution is:

[tex]\[ x = 3 \][/tex]

Therefore, Ana will hit the water 3 seconds after diving.

[tex]\[ \boxed{3} \][/tex]