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Kipish [7]
3 years ago
10

The table represents the height of a rock that is dropped, h(t), in meters, after t seconds.

Mathematics
2 answers:
laiz [17]3 years ago
5 0

Answer:

<h2>The rock hits the ground between 2 and 2.5 seconds.</h2>

Step-by-step explanation:

The given table is:

t            h(t)

0           20

0.5        18.8

1             15.1

1.5          9

2            0.4

2.5        -10.6

3           -24.1

The ground level is when h(t) = 0, which is between 0.4 and -10.6, because in that interval is included the zero. So, the problem is asking when does the rock hit the ground, referring to time. We just have to look for t-values for 0.4 and -10.6.

<em>Therefore, the rock hits the ground between 2 and 2.5 seconds.</em>

harina [27]3 years ago
4 0

Answer:

The rock hits the ground between 2

seconds and 2.5

seconds after it is dropped.

Step-by-step explanation:

At 2 seconds, the height of the rock is 0.4 meters, and at 2.5 seconds the height of the rock is -10.6 meters. Therefore, the rock hit the ground, or 0 meters, between those times.

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Answer:

Down payment

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Step-by-step explanation:

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3 0
2 years ago
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
2 years ago
What is 763 thousandths in scientific notation?
irina1246 [14]
7.63 ^-10!!!!!!!!!!!!!!!!!!!!!!!!

4 0
3 years ago
Pls solve this plèeeeeeeeease
gtnhenbr [62]

Answer:

\dfrac{5^7}{2} = \dfrac{78125}{2}

Step-by-step explanation:

\left( \dfrac{4}{5} \right)^2 \times 5^4 \times \left( \dfrac{2}{5} \right)^{-2} \div \left( \dfrac{5}{2} \right)^{-3} =

= \dfrac{4^2}{5^2} \times 5^4 \times \dfrac{5^2}{2^2} \times \dfrac{5^3}{2^3}

= \dfrac{(2^2)^2}{5^2} \times 5^4 \times \dfrac{5^2}{2^2} \times \dfrac{5^3}{2^3}

= \dfrac{2^4}{5^2} \times 5^4 \times \dfrac{5^2}{2^2} \times \dfrac{5^3}{2^3}

= 2^{4 - 2 - 3} \times 5^{- 2 + 4 + 2 + 3}

= 2^{-1} \times 5^{7}

= \dfrac{5^7}{2}

= \dfrac{78125}{2}

3 0
3 years ago
Can you please answer the question ty.
Ne4ueva [31]
Put the equation on cymath and it should help...
4 0
2 years ago
Read 2 more answers
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