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makvit [3.9K]
1 year ago
14

a big water tank contained 200 litres of water. during the day, 88 l 580 ml of water was used and 200 ml leaked. how much water

was left in the water tank?
Mathematics
1 answer:
algol131 year ago
3 0

The quantity of water left in the tank as described in the task content is; 111 l 220ml.

<h3>What is the amount of water left in the tank?</h3>

The tank described in the task content initially contains 200 liters, in which case, 1 liter corresponds to 1000ml.

Hence, when 88l 580ml was used, the remaining quantity of water is; 200l - (88l 580ml) = 111 l 420ml.

And finally, since 200 ml leaked, the quantity of water left is; 111l 220ml.

Read more on subtraction;

brainly.com/question/144291

#SPJ1

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Which of the following expressions is equivalent to (8x)^2/3 ?
miv72 [106K]

Answer:

<h3>4x^2/3</h3>

Step-by-step explanation:

Given the expression (8x)^2/3

This can also be expressed as;

(8x)^2/3

= (2^3x)^2/3

= (2³)^2/3 * x^2/3

= 2^2 * x^2/3

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3x+14=11 <br> a. 4<br> b.-1<br> c. 3
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3 years ago
At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m
beks73 [17]
<h2>Answer:</h2>

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=4.9t^{2}

and,

v_{x}=1.39a_{x}+2.5

<h2>Step-by-step explanation:</h2>

In the question,

Taking the elevation of pool along the y-axis, and length of the board along the x-axis.

On drawing the illustration in the co-ordinate system we get,

lₓ = 2 m

uₓ = 2.5 m/s

and,

h_{y}=9.5\,m

So,

From the equations of the laws of motion we can state that,

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}

So,

On putting the values we can say that,

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=(0)t+\frac{1}{2}(9.8)t^{2}\\t^{2}=\frac{9.5}{4.9}\\t^{2}=1.93\\t=1.39\,s

Now,

The <u>equation of the motion in the horizontal</u> can be given by,

v_{x}=u_{x}+a_{x}t\\v_{x}=2.5+a_{x}(1.39)\\So,\\v_{x}=1.39a_{x}+2.5

<em><u>Therefore, the equations of the motions in the horizontal and verticals are,</u></em>

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=4.9t^{2}

and,

v_{x}=1.39a_{x}+2.5

6 0
3 years ago
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