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Sergeu [11.5K]
3 years ago
5

The value of $\sqrt{73}$ is between two positive, consecutive integers. What is the product of these two integers?

Mathematics
1 answer:
AnnyKZ [126]3 years ago
3 0

Answer:

Step-by-step explanation:

What are the two closest square numbers to 73?

64 and 81

\sqrt{64} \leq \sqrt{73} \leq  \sqrt{81}\\8\leq \sqrt{73} \leq 9\\8*9 = 72

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14.44*6=86.64
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3. Same thing as number one. Find the face of one side:
1 1/3*1 1/3 is about 1.7777
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4. For this one, you want to follow the surface area formula for a cylinder. Which is - A=2πrh+2πr^2
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3 years ago
The product of two rational numbers is -25/18. If one of the numbers is -3/7, find the other
MissTica
Hello,

-3/7*x=-25/18
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What is the period of y=4cos5x
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What is a quick and easy way to remember explicit and recursive formulas?
Oliga [24]
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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

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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,

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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}
a_{n+1}=r^3(ra_{n-3})
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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
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