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2 years ago

10

IM USING UP SO MANY POINTS CAUSE PEOPLE WONT ANSWER PLZ HELP-A business owes $8000 on a loan. Every month, the business pays 12

of the amount remaining on the loan.
How much will the business pay in the sixth month?

1 answer:

allsm [11]2 years ago

8 0

He pays $125 during the sixth month.

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What is the value of f(g(3))

denpristay [2]

Answer:

A

Step-by-step explanation:

f(g(3)= f(g(x=3)) = f( $\frac{3+1}{3-1}$ ) = f(4/2) = f(2) = f(x=2) = 3*2-7 = 6-7 = -1

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2 years ago

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4 16/17 as a improper fraction

Allushta [10]

Answer:

$\frac{84}{17}$

Step-by-step explanation:

Formula for mixed number to improper fraction: (Whole number × denominator) + numerator
4 × 17 = 68, 68 + 16 = 84
Re-write the fraction: $\frac{84}{17}$

I hope this helps!

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3 years ago

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Round 71290 to the nearest 10 what's the answer?

Aloiza [94]

71290 because the ones are 0 and zero is lower than 5
so it wouldn't add anything to the tens

the rule is if the number is more than 5+ it will add 1 to the next number
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6 0

3 years ago

the constant of proportionality is always the point_______, where k is the constant of proportionality. Additionally, you can fi

MArishka [77]

Answer:

Step-by-step explanation:

When two variables say x and y are proportional let us assume y dependent variable and x independent variable

then we have y =kx

Here k is called the constant of proportionality.

Whenever x increases/decreases by 1 unit, the y value also increases/decreases by k units.

Whenever x=1, y =k

and always $\frac{y}{x} =k$

Thus we can fill up as

the constant of proportionality is always the point___(1.k)____, where k is the constant of proportionality. Additionally, you can find the constant of proportionality by finding the ratio of___y to x____, for any point on the___graph of the function.___.

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2 years ago

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

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2 years ago