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ElenaW [278]
3 years ago
14

Answer this please!!! Need help now!

Mathematics
1 answer:
Simora [160]3 years ago
6 0
It would be D...
not quite sure, cause most of these are not a function
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(urgent please help!!!!)in computing the mean of 8 numbers, a student mistakenly used 17 instead of 25 as one of the numbers and
mamaluj [8]
I believe the correct mean is 21!
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3 years ago
Read 2 more answers
Can someone help me please
laiz [17]

Area of trapezium= 1/2 × (a+b) × h

= 1/2 × (12 + 8) × 8

= 1/2 × 20 × 8

= 10 × 8

= 80 cm^3

Hope it helps you

3 0
2 years ago
Consider the function H(x) = (x ^ 2 - 3x - 4)/(x - 4)
tigry1 [53]

Answer:

this is probably the answer lololol

3 0
3 years ago
Is -4x + y = -9 parallel to x + 4y = 12
expeople1 [14]

Answer:

YES

Step-by-step explanation:

5 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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