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Sholpan [36]
3 years ago
8

A pie is 3/4 and 2/3 of the pie is eaten what is left

Mathematics
1 answer:
Sidana [21]3 years ago
4 0
Answer:
1/12 is left

Step by step explanation:
Find the common denominator which is 12 then you would get 9/12 (Originally 3/4) and 8/12 (Originally 2/3) and then you subtract which gets you 1/12
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Thrush us a landscape architect. For his first public project he is asked a small scale drawing of a garden to be placed in the
ehidna [41]

Answer:

Therefore, on the graph;

The height = 7.5 units

And base = 5.0 units

Attached is an image for further information;

Step-by-step explanation:

Given that;

1 unit on the grid represent 2m of the garden;

Ratio = 1/2 unit/m

For the height;

height h = 15m

On the graph;

h = 15m × 1/2 unit/m

h = 7.5 units

For the base;

Base b = 10m

On the graph;

b = 10m × 1/2 unit/m

b = 5 units

Therefore, on the graph;

The height = 7.5 units

And base = 5.0 units

6 0
3 years ago
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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