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daser333 [38]
3 years ago
5

An object is launched from a platform. Its height (in meters), xxx seconds after the launch, is modeled by h(x)=-5(x+1)(x-9)h(x)

=−5(x+1)(x−9)h, left parenthesis, x, right parenthesis, equals, minus, 5, left parenthesis, x, plus, 1, right parenthesis, left parenthesis, x, minus, 9, right parenthesis How many seconds after launch will the object hit the ground?
Mathematics
2 answers:
Colt1911 [192]3 years ago
4 0

Answer:

IT'S 9 SECONDS!!!!

Step-by-step explanation:

i got it wrong by 45 and when i put in 9 it was right

larisa [96]3 years ago
3 0

Answer:

The object will hit the ground after <u>9 seconds</u>.

Step-by-step explanation:

Given:

The height 'h' of an object varies with time 'x' as:

h(x)=-5(x+1)(x-9)

Now, we are asked to find the time 'x' when the height reaches ground after launch.

So, the height of the object on reaching ground will be 0 m. So, substituting 0 for 'h' and solving for 'x', we get:

0=-5(x+1)(x-9)\\\\\frac{0}{-5}=(x+1)(x-9)\\\\(x+1)(x-9)=0\\\\x+1=0\ or\ x-9=0\\\\x=-1\ or\ x=9

Therefore, there are two values of 'x' for which height is 0. The negative value is ignored as time can't be negative.

So, the value of 'x' is 9 seconds.

Hence, the object will hit the ground after 9 seconds.

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