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leonid [27]
2 years ago
10

A trough is 10 ft long and its ends have the shape of isosceles triangles that are 4 ft across at the top and have a height of 1

ft. If the trough is being filled with water at a rate of 15 ft3/min, how fast is the water level rising when the water is 8 inches deep
Mathematics
1 answer:
Anna11 [10]2 years ago
7 0

Answer:

7.5 ft³/min

Step-by-step explanation:

Let x be the depth below the surface of the water. The height, h of the water is thus h = 10 - x.

Now, the volume of water V = Ah where A = area of isosceles base of trough = 1/2bh' where b = base of triangle = 4 ft and h' = height of triangle = 1 ft. So, A = 1/2 × 4 ft × 1 ft = 2 ft²

So, V = Ah = 2h = 2(10 - x)

The rate of change of volume is thus

dV/dt = d[2(10 - x)]/dt = -2dx/dt

Since dV/dt = 15 ft³/min,

dx/dt = -(dV/dt)/2 = -15 ft³/min ÷ 2 = -7.5 ft³/min

Since the height of the water is h = 10 - x, the rate at which the water level is rising is dh/dt = d[10 - x]/dt

= -dx/dt

= -(-7.5 ft³/min)

= 7.5 ft³/min

And the height at this point when x = 8 inches = 8 in × 1 ft/12 in  = 0.67 ft is h = (10 - 0.67) ft = 9.33 ft

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Are two ratios are equivalent to two thirds
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2:3 and 4:6 are two different ratios that are equivalent to 2/3.

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3 years ago
A survey of 500 high school students was taken to determine their favorite chocolate candy. Of the 500 students surveyed, 42 lik
Step2247 [10]

Answer:

Step-by-step explanation:

Universal set

U = 500

The number that likes snickers

n(S) = 42

The number that like Twix.

n(T) = 110

The number that like Reeses

n(R) = 125

n(S n T) = 33

n(T n R) = 62

n(S n R) = 26

n( S n R n T) = 22

Then,

n(S n T) only = n(S n T) - n(S n R n T)

n(S n T) only =33 - 22 = 11

n(T n R) only = n(T n R) - n(S n R n T)

n(T n R) only =62 - 22 = 40

n(S n R) only = n(S n R) - n(S n R n T)

n(S n R) only =26 - 22 = 4.

Also,

n(S) only = n(S) - n(S n R) - n(S n T) only

n(S) only = 42 - 26 - 11 = 5

n(T) only = n(T) - n(T n R) - n(T n S) only

n(T) only = 110 - 62 - 11 = 37

n(R) only = n(R) - n(S n R) - n(R n T) only

n(R) only = 124 - 26 - 40 = 58

Then, to know if some student don't like any of the of chocolate,

Let know the number of students that like the chocolate candy

n(S) + n(T)only + n(T n R)only + n(R) only

42 + 37 + 40 + 58 = 177 students.

Therefore, the total students that like chocolate candy is 177, so those  that does not like any of them are

n(S U R U T)' = U – n(S U R U T)

n(S U R U T) = 500 - 177 = 323.

So, the question is how many student likes at most 2 kinds of these chocolates, this means that they can like exactly 2 or less or even none.

So, this category are

n(2 most) = n(S n T)only + n(R n T)only + n(S n R)only + n(s)only + n(T)only + n(R)only + n(S U R U T)'

n(2 most) = 11 + 40 + 4 + 5 + 37 + 58 + 323

n(2 most) = 478

The correct answer is E.

Check attachment for Venn diagram

5 0
3 years ago
what digit is in the ten thousands place 1.03098 which is the correct answer I know it's one of the zeros I believe which one is
elena-14-01-66 [18.8K]
Hello there

Actually, the correct answer is the 9

the first 0 is in the tenths spot, the 3 is in the hundredths, the second zero is in the thousandths, and the 9 is in the ten thousandths, while the 8 is in the hundred thousandths.

Therefore, the 9 is in the ten thousandths place

I hope this helped ^^ 
4 0
3 years ago
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