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Dima020 [189]
3 years ago
10

How many 1/4 litre glasses of con be got from a container of 3 litres​

Mathematics
1 answer:
Klio2033 [76]3 years ago
7 0
Total volume = 3 litres
Volume of each glass = 1/4 litre

Divide 3 by 1/4
3 / (1/4)
3* (4/1)
= 12 glasses
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Which rational number equals 0 point 4 with bar over 4?
Ulleksa [173]
<h3>Answer:   4/9</h3>

==============================================

Work Shown:

x = 0.\overline{4}\\\\x = 0.4444\ldots\\\\10x = 4.4444\ldots\\\\10x-x = 4.4444\ldots-0.4444\ldots\\\\9x = 4\\\\x = \frac{4}{9}\\\\\frac{4}{9} = 0.\overline{4}= 0.4444\ldots\\\\

The idea is that when we subtract 10x and x, the infinite decimal pattern (the 4s that go on forever) will cancel. That leads to 9x = 4 which solves to x = 4/9

6 0
3 years ago
Decrease £390 by 10%
Savatey [412]

Answer:

the other guy is correct, i think

Step-by-step explanation:

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2 years ago
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Reil [10]
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8 0
3 years ago
You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along
sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}

Then, the time (speed divided by distance) is:

t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1

To optimize this function we have to derive and equal to zero:

\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\  \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\

x

As d_b=x, the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

7 0
3 years ago
Can someone please help me with this question? I have been stuck on it for a while now​
frez [133]

Answer:

<em>w = 12 cm  </em>and <em> l = 20 cm</em>

Step-by-step explanation:

3 0
3 years ago
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