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3 years ago

11

A cylindrical water tank has a diameter of 4 m and a height of 3 m, while a water tank shaped like a rectangular prism has a len

gth of 7 m, a width of 2 meters and a height of 2 meters. Which of the two tanks can hold more water? By how many cubic meters

1 answer:

katrin2010 [14]3 years ago

7 0

The 1st one can vary more by 12 cubic meters

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3 years ago

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Is 6 (7) + 2 (3²) = 60?

sertanlavr [38]

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Step-by-step explanation:

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3 years ago

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Find the midpoint of the line segment whose endpoints are (0,1/2 ) and (0,3/4 ).

Ksivusya [100]

Formula of the midpoint:
x =(x₁+x₂)/2 and y = (y₁+y₂)/2

Plugin the respective value
for A(0,1/2) and B(0, 3/4)

x= (0+0)/2 = 0
y = (1/2 + 3/4)/2 = 5/8

Then the midpoint of the line segment whose endpoints are (0,1/2 ) and (0,3/4 ). is M(0, 5/8) or M(0, 0.625)

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4 years ago

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Find the sum Sn below:

Hoochie [10]

We can write <em>S</em> as

$\displaystyle S = \sum_{k=0}^{n-1} (n-k)3^k$

and expand it as

$\displaystyle S = n \sum_{k=0}^{n-1} 3^k - \sum_{k=0}^{n-1} k\cdot3^k$

The first sum is geometric, nothing tricky:

$\displaystyle\sum_{k=0}^{n-1} 3^k = 1 + 3 + 3^2 + \cdots + 3^{n-1} \\\\ \implies 3\sum_{k=0}^{n-1} 3^k = 3 + 3^2 + 3^3 + \cdots + 3^n \\\\ \implies -2\sum_{k=0}^{n-1} 3^k = 1 - 3^n \\\\ \implies \sum_{k=0}^{n-1} 3^k = \frac{3^n-1}2$

For the second sum, you can use the same method employed in another question of yours (24494877) to find

$\displaystyle \sum_{k=0}^{n-1} k\cdot 3^k = \frac{(2n-3)3^n+3}4$

So this sum comes out to

$\displaystyle S = n\cdot\frac{3^n-1}2 - \frac{(2n-3)3^n+3}4 \\\\ S = \frac{2n\cdot3^n-2n - (2n-3)3^n-3}4 \\\\ \boxed{S = \frac{3^{n+1}-2n-3}4}$

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3 years ago