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daser333 [38]
4 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]4 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]4 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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stiks02 [169]
4 fits into 24 6 times so 24/4 written as a whole number is 6! I hope this helped :D
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Which of the following best describes the equation below?
sergiy2304 [10]

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both linear and nonlinear

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A cubic box is completely filled with 2,179 g of water. What is the length of one side of the box, in meters? m Explain your rea
Mazyrski [523]

Answer:

l\approx 1.296\,cm

Step-by-step explanation:

The water density is approximately 1 gram per cubic centimeter, which means that 1 gram of water occupies a volume of a cubic centimeter. Hence, 2.179 grams occupies a volume of 2.179 cubic centimeters.

Let assume volume occupied by the water has a cube-like shape, whose side has the following measure:

l = \sqrt[3]{2.179\,m^{3}}

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5 0
3 years ago
Add: (–21) + (–37) + (–15)
Rina8888 [55]

Answer:

the answer is -73

Step-by-step explanation:


4 0
3 years ago
A local gym instructor has a course load that allows her to teach eight classes. At an interest meeting, 8 people wanted high-­‐
cricket20 [7]

Answer:

<u>The final curse load of the local gym instructor is this:</u>

<u>High-­‐impact aerobics = 1</u>

<u>Low-­‐impact aerobics =  4</u>

<u>Jazzercise = 1</u>

<u>Step exercise = 2</u>

Step-by-step explanation:

1. Let's apportion the class load using the Hamilton method

Number of classes the local gym instructor can teach = 8

Total number of students that want to take a class = 114 (8 people wanted high-­‐impact aerobics, 64 wanted low-­‐impact aerobics, 11 wanted jazzercise, and 31 wanted step exercise)

Standard divisor = 114/8 = 14.25

Now, we can apportion the students in in Standard quotas, this way:

High-­‐impact aerobics = 8/14.25 = 0.5614

Low-­‐impact aerobics = 64/14.25 = 4.49

Jazzercise = 11/14.25 = 0.7719

Step exercise = 31/14.25 = 2.1754

Now, we find the Minimum quota, just considering the whole number and don't taking into account the decimals, this way:

High-­‐impact aerobics = 0

Low-­‐impact aerobics =  4

Jazzercise = 0

Step exercise = 2

As we can see we have 6 classes and there are 2 still pending. Those 2 goes to the classes with the highest decimal portion, in this case, Jazzercise .7719 and High-­‐impact aerobics with .5614.

<u>The final course load is this:</u>

<u>High-­‐impact aerobics = 1</u>

<u>Low-­‐impact aerobics =  4</u>

<u>Jazzercise = 1</u>

<u>Step exercise = 2</u>

7 0
3 years ago
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