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STALIN [3.7K]
3 years ago
8

For which of the following is x = -5 a solution? Select all that apply.

Mathematics
1 answer:
pantera1 [17]3 years ago
7 0

Answer:

-x2 + 50 = 25

5x<0

2x <10

Step-by-step explanation:

I hope the answer satisfies you

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ASAP 55 points Which expression is equivalent to ((x^(-4)y)/(x^(-9)y^(5)))^(-2) assume x = 0 ,y = 0
german

Answer:

I thought if you try to do an exponent on a zero, then the answer would always be undefined...

Step-by-step explanation:

6 0
3 years ago
7g + 11 subtracted from 12
BlackZzzverrR [31]
A subtracted from b means b-a

7g+11 subtracted from 12 means 12-(7g+11)=12-1(7g+11)=12-7g-11=12-11-7g=1-7g
5 0
2 years ago
What is the answer to q divided by 1/3= 1 1/2
katrin2010 [14]
The answer is q = 3/6.
7 0
2 years ago
Read 2 more answers
Mr. Ortiz pays a state tax of 10.4% on certain purchases. Which expression is the same as the total cost of p, a purchase of $14
Nata [24]

Answer:

idk

Step-by-step explanation:

IDK

4 0
3 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
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
3 years ago
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