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Gelneren [198K]
3 years ago
7

What is 54 divided by 6

Mathematics
1 answer:
kap26 [50]3 years ago
5 0

Answer:

43

Step-by-step explanation:

You might be interested in
Keith won a cash prize in an art competition. He gave his mother 2/5 of the money he won. Of the remainder, 1/4 was given to his
USPshnik [31]

Answer:

300 dollars

Step-by-step explanation:

Let the original amount = x

x - 2/5x  was what his mother's share looked.

5/5 x - 2/5 x

3/5x is what is left.

3/5 x - 1/4 * 3/5 x - 2/3x * 3/5 = his sister's share and his brothers

3/5x - 1/4*3/5 x - 2/3 * 3/5 = 15

3/5x * (1 - 1/4 - 2/3) = 15

3/5 x 5/60 = 15

15 / 300 * x = 15

1/20  x = 15

x = 15 * 20

x = 300

7 0
2 years ago
Find the sum of (x+5), (-4x-2), and (2x-1)
RideAnS [48]
I think the answer is -x+2.
3 0
3 years ago
What should be the first step in adding these equations to eliminate y?
ElenaW [278]

Answer:

D

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
2,17,82,257,626,1297 next one please ?​
In-s [12.5K]

The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule n^4+1. The next number would then be fourth power of 7 plus 1, or 2402.

And the harder way: Denote the <em>n</em>-th term in this sequence by a_n, and denote the given sequence by \{a_n\}_{n\ge1}.

Let b_n denote the <em>n</em>-th term in the sequence of forward differences of \{a_n\}, defined by

b_n=a_{n+1}-a_n

for <em>n</em> ≥ 1. That is, \{b_n\} is the sequence with

b_1=a_2-a_1=17-2=15

b_2=a_3-a_2=82-17=65

b_3=a_4-a_3=175

b_4=a_5-a_4=369

b_5=a_6-a_5=671

and so on.

Next, let c_n denote the <em>n</em>-th term of the differences of \{b_n\}, i.e. for <em>n</em> ≥ 1,

c_n=b_{n+1}-b_n

so that

c_1=b_2-b_1=65-15=50

c_2=110

c_3=194

c_4=302

etc.

Again: let d_n denote the <em>n</em>-th difference of \{c_n\}:

d_n=c_{n+1}-c_n

d_1=c_2-c_1=60

d_2=84

d_3=108

etc.

One more time: let e_n denote the <em>n</em>-th difference of \{d_n\}:

e_n=d_{n+1}-d_n

e_1=d_2-d_1=24

e_2=24

etc.

The fact that these last differences are constant is a good sign that e_n=24 for all <em>n</em> ≥ 1. Assuming this, we would see that \{d_n\} is an arithmetic sequence given recursively by

\begin{cases}d_1=60\\d_{n+1}=d_n+24&\text{for }n>1\end{cases}

and we can easily find the explicit rule:

d_2=d_1+24

d_3=d_2+24=d_1+24\cdot2

d_4=d_3+24=d_1+24\cdot3

and so on, up to

d_n=d_1+24(n-1)

d_n=24n+36

Use the same strategy to find a closed form for \{c_n\}, then for \{b_n\}, and finally \{a_n\}.

\begin{cases}c_1=50\\c_{n+1}=c_n+24n+36&\text{for }n>1\end{cases}

c_2=c_1+24\cdot1+36

c_3=c_2+24\cdot2+36=c_1+24(1+2)+36\cdot2

c_4=c_3+24\cdot3+36=c_1+24(1+2+3)+36\cdot3

and so on, up to

c_n=c_1+24(1+2+3+\cdots+(n-1))+36(n-1)

Recall the formula for the sum of consecutive integers:

1+2+3+\cdots+n=\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2

\implies c_n=c_1+\dfrac{24(n-1)n}2+36(n-1)

\implies c_n=12n^2+24n+14

\begin{cases}b_1=15\\b_{n+1}=b_n+12n^2+24n+14&\text{for }n>1\end{cases}

b_2=b_1+12\cdot1^2+24\cdot1+14

b_3=b_2+12\cdot2^2+24\cdot2+14=b_1+12(1^2+2^2)+24(1+2)+14\cdot2

b_4=b_3+12\cdot3^2+24\cdot3+14=b_1+12(1^2+2^2+3^2)+24(1+2+3)+14\cdot3

and so on, up to

b_n=b_1+12(1^2+2^2+3^2+\cdots+(n-1)^2)+24(1+2+3+\cdots+(n-1))+14(n-1)

Recall the formula for the sum of squares of consecutive integers:

1^2+2^2+3^2+\cdots+n^2=\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6

\implies b_n=15+\dfrac{12(n-1)n(2(n-1)+1)}6+\dfrac{24(n-1)n}2+14(n-1)

\implies b_n=4n^3+6n^2+4n+1

\begin{cases}a_1=2\\a_{n+1}=a_n+4n^3+6n^2+4n+1&\text{for }n>1\end{cases}

a_2=a_1+4\cdot1^3+6\cdot1^2+4\cdot1+1

a_3=a_2+4(1^3+2^3)+6(1^2+2^2)+4(1+2)+1\cdot2

a_4=a_3+4(1^3+2^3+3^3)+6(1^2+2^2+3^2)+4(1+2+3)+1\cdot3

\implies a_n=a_1+4\displaystyle\sum_{k=1}^3k^3+6\sum_{k=1}^3k^2+4\sum_{k=1}^3k+\sum_{k=1}^{n-1}1

\displaystyle\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4

\implies a_n=2+\dfrac{4(n-1)^2n^2}4+\dfrac{6(n-1)n(2n)}6+\dfrac{4(n-1)n}2+(n-1)

\implies a_n=n^4+1

4 0
3 years ago
Fiona invested $1000 at 7% compounded continuously. At the same time, Maria invested $1100 at 7% compounded daily. How long will
Natali [406]

9514 1404 393

Answer:

  14,201 years

Step-by-step explanation:

The two compound interest formulas are ...

  A = P·e^(rt) . . . . . continuous compounding at rate r for t years

  A = P·(1 +r/365)^(365t) . . . . . daily compounding at rate r for t years

We went the amounts to be equal:

  1000·e^(0.07t) = 1100·(1+0.07/365)^(365t)

Dividing by 1000(1 +0.07/365)^(365t), we have ...

  ((e^0.07)/(1+0.07/365)^365)^t = 1.1

The base of the exponential on the left is ...

 ( e^0.07)/(1+0.07/365)^365 ≈ 1.00000671149321522

Taking logs, we have ...

  t×ln(1.00000671149321522) = ln(1.1)

  t = ln(1.1)/ln(1.00000671149321522) ≈ 0.09531018/(6.7114704·10^-6)

  t ≈ 14,201.09 . . . . . years

It will take about 14,201 years for the investments to be equal.

_____

<em>Additional comment</em>

The investment value at that time will be about $5.269·10^434. (That's a larger number than <em>anything</em> countable in the known universe, including energy quanta.)

These calculations are beyond the ability of many calculators, so might need to be carefully rewritten if the calculator only keeps 10 significant digits, or only manages exponents less than 100.

This shows that daily compounding is very close in effect to continuous compounding. It would take almost 150 years to make a difference of 0.1% in value.

4 0
3 years ago
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