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3 years ago

What are some examples from science and engineering where a quadratic is used but is not a function?

Mathematics

1 answer:

dem82 [27]3 years ago

7 0

Answer:

Quadratic equations are important when designing curved equipment such as auto-bodies. Brake systems are designed by automotive engineers by solving equations that arise. Aerospace engineers also interact with quadratic equations so often in their careers.

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A store hands out 900 reward cards. Customers scan their cards in the store to find out what reward they receive. The probabilit

natima [27]

Answer:

75 customers

Step-by-step explanation: Given that

there are 900 gift card

1/12 of them are worth $25,

then the number of customers that received $25 will be 1/12 of 900. That is

1/12 × 900 = 75 customers

5 0

4 years ago

Which expression has a value of 2/3?

Bas_tet [7]

Answer:

You need to add an attachment or type the answer choices.

Step-by-step explanation:

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3 years ago

Is the following table a function? Why?

KonstantinChe [14]

Answer:

No, because on the Y row, it has both of the same numbers on that one row. Too make it more simpler, The y row has both 19 on it. Therefore, it is not a function.

Step-by-step explanation:

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3 0

3 years ago

A potato was launched in the air using a potato gun. The function for the situation was h(t)=-16t^2 +100t+10, where t was the ti

Goshia [24]

Answer:

The height equation is:

h(t)=-16t^2 +100t+10

A) To find the maximum height, we must find the time where the velocity is 0, so we must derive it with respect to the time.

v(t) = -2*16*t + 100 = 0

t = 100/(2*16) = 3.125s

Now we put this time in our height equation:

h(3.125s) = -16*3.125^2 + 100*3.125 + 10 = 166.25 ft.

B) The lowest heigth of the potato will be h = 0ft, when the potato hits the ground.

C) the lowest time that works for that function is t = 0s, when the potato is fired by the gun.

D) the maximum time can be finded when the potato hits the ground, afther that point the equation does not work anymore, let's find it,

h(t) = 0 = -16t^2 +100t+10

we can solve it using Bhaskara's equation:

$t = \frac{-100 +- \sqrt{100^2 +4*16*10} }{-2*16} = \frac{-100 +- 103.2}{-32}$

So the two solutions are:

t = 6.3s

t = -0.1s

We need to chose the positive time, as we already discused that the minimum time that works for the equation is t = 0s.

so here the answer is t = 6.3s

E) the domain is 0s ≤ t ≤ 6.3s

The range is 0ft ≤ h ≤ 166.25 ft.

4 0

3 years ago

Solve the inequality and express your answer in interval notation. x^2- 8x+3<0

Papessa [141]

Solution:

A. $(-4-\sqrt{13},-4+\sqrt{13})$

Hope this helps!

7 0

2 years ago