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Mashutka [201]
3 years ago
8

Need help ASAP please thanks.

Mathematics
1 answer:
Rus_ich [418]3 years ago
6 0

Answer: 13,832

Step-by-step explanation:

18^3 power is 5,832 and 20^3 power is 8,000

add them together, you would get 13,832

I would go with D

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I believe the correct answer is 2,300,500,004
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A flow meter is attached in a hydraulic line that measures 18 gal/min. The line has an inside diameter of 1.0 in. What is the fl
snow_tiger [21]

The flow velocity measured as in/min at the meter is 5924.13 in/min

<h3 /><h3>Volumetric flow rate</h3>

We know that the volume flow rate Q = Av where

  • A = cross-sectional area of pipe and
  • v = flow velocity

Now, Q = 18 gal/min

Converting this to in³/min, we have

Q = 18 gal/min = 18 × 1 gal/min = 18 × 231 in³/min = 4158 in³/min

A = πd²/4 where d = diameter of pipe = 1.0 in.

<h3 /><h3>Flow velocity, v</h3>

Since Q = Av, making v subject of the formula, we have

v = Q/A

v = 4Q/πd²

Substituting the values of the variables into the equation, we have

v = 4Q/πd²

v = 4 × 4158 in³/min ÷ π × (1.0 in)²

v = 16632 in/min ÷ π

v = 5924.13 in/min

So, the flow velocity measured as in/min at the meter is 5924.13 in/min

Learn more about flow velocity here:

brainly.com/question/15648466

5 0
2 years ago
Your car invoice reads: “Price $1,492.50, dealer preparation $45.00, transportation $54.50, undercoat $148.80, 60-day guarantee
Maurinko [17]

Answer:$2,074.53

Step-by-step explanation:

Final cost will be calculated as the sum of each expense stated on the invoice, which is $1,492.50 + $45 + $54.50+ $148.80 + $95 + $238.73 = $2,074.53

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7 0
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Answer: Sorry this took so long, here is the photo

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6 0
2 years ago
What is a quick and easy way to remember explicit and recursive formulas?
Oliga [24]
I always found derivation to be helpful in remembering. Since this question is tagged as at the middle school level, I assume you've only learned about arithmetic and geometric sequences.

First, remember what these names mean. An arithmetic sequence is a sequence in which consecutive terms are increased by a fixed amount; in other words, it is an additive sequence. If a_n is the nth term in the sequence, then the next term a_{n+1} is a fixed constant (the common difference d) added to the previous term. As a recursive formula, that's

a_{n+1}=a_n+d

This is the part that's probably easier for you to remember. The explicit formula is easily derived from this definition. Since a_{n+1}=a_n+d, this means that a_n=a_{n-1}+d, so you plug this into the recursive formula and end up with 

a_{n+1}=(a_{n-1}+d)+d=a_{n-1}+2d

You can continue in this pattern, since every term in the sequence follows this rule:

a_{n+1}=a_{n-1}+2d
a_{n+1}=(a_{n-2}+d)+2d
a_{n+1}=a_{n-2}+3d
a_{n+1}=(a_{n-3}+d)+3d
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and so on. You start to notice a pattern: the subscript of the earlier term in the sequence (on the right side) and the coefficient of the common difference always add up to n+1. You have, for example, (n-2)+3=n+1 in the third equation above.

Continuing this pattern, you can write the formula in terms of a known number in the sequence, typically the first one a_1. In order for the pattern mentioned above to hold, you would end up with

a_{n+1}=a_1+nd

or, shifting the index by one so that the formula gives the nth term explicitly,

a_n=a_1+(n-1)d

Now, geometric sequences behave similarly, but instead of changing additively, the terms of the sequence are scaled or changed multiplicatively. In other words, there is some fixed common ratio r between terms that scales the next term in the sequence relative to the previous one. As a recursive formula,

a_{n+1}=ra_n

Well, since a_n is just the term after a_{n-1} scaled by r, you can write

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Doing this again and again, you'll see a similar pattern emerge:

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a_{n+1}=r^3a_{n-2}
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a_{n+1}=r^4a_{n-3}

and so on. Notice that the subscript and the exponent of the common ratio both add up to n+1. For instance, in the third equation, 3+(n-2)=n+1. Extrapolating from this, you can write the explicit rule in terms of the first number in the sequence:

a_{n+1}=r^na_1

or, to give the formula for a_n explicitly,

a_n=r^{n-1}a_1
6 0
3 years ago
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