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Masja [62]
3 years ago
8

1.

Mathematics
1 answer:
Vladimir79 [104]3 years ago
7 0

Answer:

a) n = -3

Step-by-step explanation:

3.8 x 10^n = 0.0038

0.0038 = 38/10000 = 38/10^4 = 38 x 10^-4 = 3.8 x 10^-3

You might be interested in
Write each decimal as a fraction in lowest terms <br><br> 0.255
Vesnalui [34]

Answer:

51/200

Step-by-step explanation:

To write a decimal as a fraction, you have to make the decimal the numerator, and put a multiple of ten with that number of zeroes in the denominator:

0.255 = 255/1000

Then we simplify

51/200

6 0
1 year ago
What is 3 tenths more than 8.154 ? What is 2 and 2 tenths more than 1.425? Explain please !
Anna11 [10]
8.454 is 3 tenths more than 8.154
3.625 is 2 and 2 tenths more than 1.425

Place values after the decimal point are tenths, hundredths, thousandths, ten thousandths...so on.

So 3 tenths looks like this 0.30 ⇒ 3 is in the tenths place. When you say 3 tenths more, it is a clue that the operation to be done is addition. thus,

   8.154
<u>+ 0.30
</u>   8.454   * in addition or subtraction, the decimal points must be aligned to avoid confusion. 

2 and 2 tenths is like this 2.20 ⇒ 1st 2 is in the ones or units place and the 2nd 2 is in the tenths place. Again it states more than so addition must be done.
 
   1.425
<u>+ 2.20</u>
   3.625
<u>

</u>
6 0
3 years ago
Read 2 more answers
The lock is numbered from 0 to 49. Each combination uses three numbers in a right, left, right pattern. Find the total number of
sineoko [7]

Answer:

147

Step-by-step explanation:

8 0
3 years ago
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
Chang knows one side of a triangle is 13 cm. Which set of two sides is possible for the lengths of the other two sides of this t
natima [27]

Answer:

8cm and 2cm

I would appreciate if my answer is chosen as a brainliest answer

4 0
2 years ago
Read 2 more answers
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