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Aleks04 [339]
2 years ago
10

Is the equation true, false, or open? Explain. 20-1=1+17​

Mathematics
2 answers:
nikklg [1K]2 years ago
5 0
<h2>False</h2>

This statement is false:

20 – 1 = <em>19</em>

1 + 17 = <em>18</em>

We get:

<em>19</em> = <em>18</em>

Since <em>19 </em>doesn't equal <em>18, </em>the statement is false.

<em>Hope this helps :)</em>

Bad White [126]2 years ago
4 0

Answer:

false

Step-by-step explanation:

20 - 1 = 1 + 17​

Simplify each side.

19 = 18

Since 19 = 18 is a false statement, the equation is false.

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A class consists of 1/2 freshmen, 1/8 sophomores, and 1/8 juniors; the rest are seniors. What fraction of the class is seniors?
Firlakuza [10]

Answer:

2/8

Step-by-step explanation:

freshman=4/8

1/8 soph    1/8 jr

add all together u get 6/8

the rest is seniors so its 2/8

8 0
3 years ago
I need help with these. Im really lost and i need to get this done before school is over... help me out please!
gtnhenbr [62]
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3 years ago
Write an equivalent expression for 5(3h + 2) - 6.
Anna71 [15]

Answer:

15h + 4

Step-by-step explanation:

5(3h + 2) - 6

Distribute

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5 x 2 = 10

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15h + 10 - 6

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Hope This Helps!! :)

7 0
3 years ago
Read 2 more answers
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
a_{n+1}=a_{n-3}+4d

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

a_{n+1}=r(ra_{n-1})=r^2a_{n-1}

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

a_{n+1}=r^2a_{n-1}
a_{n+1}=r^2(ra_{n-2})
a_{n+1}=r^3a_{n-2}
a_{n+1}=r^3(ra_{n-3})
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
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It’s 4 m, I took the test and got it right
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