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sp2606 [1]
3 years ago
15

The height of water in a rectangular tank is given by h(w), where w is the volume of water (in liters) that has been removed fro

m the tank. The height of the water decreases 10mm per liter removed.
a. From the time when 7 L have been removed to the time when 15 L have been removed, the accumulated change in the height of the water is
b. If the initial height of the water is 600mm, then the height of the water after w liters have been removed is
Mathematics
1 answer:
zmey [24]3 years ago
7 0

Answer:

a) ΔL = 80mm

b) h = 600 - 10w mm

Step-by-step explanation:

a. From the time when 7 L have been removed to the time when 15 L have been removed, the change in the volume has been 8L and how the height of the water decreases 10mm per liter removed, then the acumulate change in the height of water has been (8lt)(10mm/lt) = 80 mm

b) If w liters have been removed, then the change in the height of water is (10mm/lt)(wlt) = 10w mm and thus the water's height at this precise point in time is 600 - 10w mm

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GaryK [48]

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Divide through by \cos \theta

r=\frac{2}{\cos \theta}

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\boxed{x=2\to r=2\sec \theta}

2. The given rectangular equation is:

x^2+y^2=36

This is the same as:

x^2+y^2=6^2

We use the relation r^2=x^2+y^2

This implies that:

r^2=6^2

\therefore r=6

\boxed{x^2+y^2=36\to r=6}

3. The given rectangular equation is:

x^2+y^2=2y

This is the same as:

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This implies that:

r^2=2r\sin \theta

Divide through by r

r=2\sin \theta

\boxed{x^2+y^2=2y\to r=2\sin \theta}

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We substitute y=r\sin \theta and x=r\cos \theta

r\cos \theta=r\sin \theta\sqrt{3}

This implies that;

\tan \theta=\frac{\sqrt{3}}{3}

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r\cos \theta=r\sin \theta

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6 0
3 years ago
Answer this thanks!!!!
kirill115 [55]

Answer:

B

Step-by-step explanation:

$65×6%=$3.90

$3.90+$65=$68.90

m I right

4 0
2 years ago
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Naya [18.7K]

Divide:

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zalisa [80]
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3 years ago
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Answer:

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