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lesya692 [45]
3 years ago
11

A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10

cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

Mathematics
1 answer:
just olya [345]3 years ago
3 0

Answer:

\frac{dh}{dt}≅0.286\frac{ft^{3} }{min}

Step-by-step explanation:

V=\frac{\pi }{3}r^{2}h; rate of change \frac{dV}{dt}=10\frac{ft^{3} }{min}, we must find the rate of change of the depth \frac{dh}{dt} =?;h=8ft

5h=12r; V=\frac{\pi }{3}\({\((5h}/12} )} ^{2}h=\frac{\pi }{3}(\frac{25h^{2} }{144})h; V=\frac{25\pi h^{3}}{432}; deriving \frac{dV}{dt} = \frac{25\pi }{432}(3h^{2})\frac{dh}{dt} → 10=\frac{25\pi h^{2}}{144} \frac{dh}{dt} → h=8 then \frac{dh}{dt}=\frac{1440}{25\pi 64}=\frac{9}{10\pi}≅ 0.286\frac{ft^{3} }{min}

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Write expression using the distributive property to find the product of 9 x82
irga5000 [103]

Answer:

Step-by-step explanation:

9 * 82 = ( 10-1) * 82

          = 10*82   - 1 *82

          = 820 - 82

           = 738

5 0
3 years ago
If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
2 years ago
PLEASE HELP THIS IS MY LAST QUSTION OF THE DAY!!! :)<br> thanks in advanced
notsponge [240]

So using standard form you convert the equation to 5x - 2y = -10 then use slope m of a line of the form Ax + By = C equals negative a over b. It would be a = 5 b = -2 m = -5/-2 m = 5/2. Have a good day the answer is 5/2

3 0
3 years ago
Change 16/20 to tenths
solniwko [45]
16/20 = 8(2)/10(2) = 8/10. 8/10 can also be further reduced to 4/5.
5 0
3 years ago
Read 2 more answers
Can someone plz help me find the GCF if number 9 plz
ivann1987 [24]
The answer should be 4, 4 can go into all of these numbers
5 0
3 years ago
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