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zhuklara [117]
3 years ago
12

I really don't know how to do this, plz help (number 11)

Mathematics
1 answer:
Pani-rosa [81]3 years ago
6 0
To know the answer for 11, we need to know the how much 1 quart equals to a liter first. One quart equals 0.94635295 liters. When we multiply this by 10, and round to the nearest hundredth we get 9.46. 
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The sum of a number and 6 is greater than or equal to –4 =
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x is greater than or = to 8

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2, 2, 3 can be lengths of the sides of a triangle. true or false
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Yes

Step-by-step explanation:

The sum of any 2 sides of a triangle are greater than the 3rd side.

2,2,3 satisfies this condition.

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A rocket is fired with an initial vertical velocity of 40 m/s from a launch pad 45 m high, and its height is given by the formul
N76 [4]

Answer:

1) 125 meters

2) For 9 seconds.

Step-by-step explanation:

The rocket's height is given by the formula -5t^2+40t+45.

Notice that this is a quadratic.

Part 1)

Since this is a quadratic, the highest height the rocket goes will be the vertex of our quadratic. Remember that the vertex of a quadratic in standard form is:

(-\frac{b}{2a}, f(-\frac{b}{2a}))

So, let's identify our coefficients. The standard quadratic form is:

ax^2+bx+c

Therefore, in this case, our a is -5, b is 40, and c is 45.

So, let's find the x-coordinate of our vertex. Substitute 40 for b and -5 for a. This yields:

x=\frac{-(40)}{2(-5)}

Multiply and reduce. So:

x=-40/-10=4

Now, substitute 4 for t to find the height. So:

-5(4)^2+40(4)+45

Evaluate:

=-5(16)+40(4)+45

Multiply and add:

=-80+160+45=125\text{ meters}

Therefore, the highest the rocket goes up is 125 meters.

Part 2)

To find out for how much time the rocket is in the air, we can think about after how many seconds after launch will the rocket land.

If the rocket lands, the height h will be 0. So, we can set our expression equal to 0 and solve for t:

-5t^2+40t+45=0

First, let's divide everything by -5. This yields:

t^2-8t-9=0

We can factor:

(t+1)(t-9)=0

Zero Product Property:

t+1=0\text{ or } t-9=0

Solve for t:

t=-1\text{ or } t=9

In this case, since t is our time in seconds, -1 seconds does not make sense. So, we can remove -1 from our solution set.

Therefore, 9 seconds after launch, the rocket will touch the ground.

Therefore, the rocket was in the air for 9 seconds.

And we're done!

6 0
3 years ago
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