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snow_tiger [21]
3 years ago
14

Michelle throws a frisbee into the air. The height of the frisbee at a given time can be modeled by the equation h(t)= -2t

Mathematics
1 answer:
viva [34]3 years ago
3 0

Answer:

(a) The ball will hit the ground after 3 seconds

(b) The maximum height is 6.125

Step-by-step explanation:

Given

h(t) = -2t^2 +5t +3

Solving (a): When the frisbee will hit the ground?

To do this, we set h(t) to 0

So, we have:

h(t) = -2t^2 +5t +3

-2t^2 +5t +3=0

Expand

-2t^2 +6t-t +3=0

Factorize

-2t(t - 3) -1(t - 3) = 0

Factor out t - 3

(-2t  -1)(t - 3) = 0

Split:

-2t  -1= 0\ or\ t - 3 = 0

Solve for t in both equations

-2t  =1\ or\ t = 3

t  =-\frac{1}{2}\ or\ t = 3

Time can't be negative; So:

t = 3

Solving (b): How height the frisbee will go?

First, we calculate time to reach the maximum height

t = -\frac{b}{2a}

Where:

h(t) = at^2 + bt + c

By comparison:

a = -2,\ b =5,\ c =3

So:

t = -\frac{b}{2a}

t = -\frac{5}{2*-2}

t = \frac{5}{4}

t = 1.25

So, the maximum height is:

h_{max} = -2 * 1.25^2 + 5 * 1.25 + 3

h_{max} = 6.125

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bogdanovich [222]

Answer:

  • \boxed{\sf{7y+3-4.5z}}

Step-by-step explanation:

In order to combine like terms, you have to isolate x and y from one side of the equation.

\sf{-8z+\left(4.5\right)+3.5z+7y-1.5}

<u>First, thing you do is remove parentheses.</u>

\Longrightarrow: \sf{-8z+4.5+3.5z+7y-1.5}

<u>Solve.</u>

<u>Then, you combine like terms.</u>

\Longrightarrow:\sf{-8z+3.5z+7y+4.5-1.5}

<u>Add/subtract the numbers from left to right.</u>

-8z+3.5z=-4.5z

<u>Rewrite the problem down.</u>

\Longrightarrow: \sf{-4.5z+7y+4.5-1.5}

<u>Solve.</u>

4.5-1.5=3

\Longrightarrow: \boxed{\sf{7y+3-4.5z}}

  • <u>Therefore, the correct answer is 7y+3-4.5z.</u>

I hope this helps, let me know if you have any questions.

6 0
1 year ago
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Answer:

5n

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cost of one helmet = $5

let 'n' be the number of person that is n=1,2,3,4........

note that the value of n is always a natural number

therefore the expression = 5n

where 5 is the cost one helmet

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5. After 6 hours, the concentration is halved. After 12 hours, it will be quartered. If the original concentration is 200 nmol, a quarter is 50 nmol.

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