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Sindrei [870]
4 years ago
13

If you are good at sequences in math please help I am giving more points

Mathematics
2 answers:
slamgirl [31]4 years ago
5 0

Answer:

-20 + 7(n-1)

Step-by-step explanation:

It's already given that d, or the common difference, is 7. The only thing left to do is to find the first term of this sequence.

We are given that a_{11} = 50.

If we subtract the common difference 10 times we can find the first term.

- 7 x 10 = -70

50 - 70 = -20

The first term is -20. The only choice that shows this is the first one.

Mrrafil [7]4 years ago
3 0

Answer:

First one

Step-by-step explanation:

a11 = a1 + (11-1)(7)

50 = a1 + 70

a1 = -20

an = -20 + 7(n-1)

You might be interested in
Solve the division problem 15/4/-5/8
artcher [175]

Answer: -6

<u>Do Keep Change Flip (KCF)</u>

Keep: 15/4

Change: ÷ into ×

Flip: -5/8 into 8/-5

<u>Multiply</u>

15/4*8/-5=120/-20

<u>Divide</u>

120/-20=-6

6 0
3 years ago
Read 2 more answers
Suppose we want to choose 5 objects, without replacement, from 13 distinct objects. (a) How many ways can this be done, if the o
Arisa [49]

Answer:

A. 1, 287 ways

B. 154,440 ways

Step-by-step explanation:

A. We want to choose 5 objects from a total 13, without considering the order in which they are chosen.

The correct way to do this is by using the combination formula since order is not considered;

Thus we have ; 13 C 5 read as 13 combination 5;

Mathematically, n C r is ; n!/(n-r)!r!

Thus, we have ;

13!/(13-8)!8! = 13!/5!8! = 1,287 ways

B. By considering order, we shall be using the permutation formula;

Mathematically n P r = n!/(n-r)!

Read as n permutation r;

Using the numbers involved, we have ; 13 P 5

= 13!/(13-5)! = 13!/8! = 154,440 ways

8 0
3 years ago
6949 * 7/8 simplify. please help?
madam [21]
The answer to this question is:

6080 3/8
7 0
3 years ago
Read 2 more answers
Help me ASAP for the brainliest answer
ikadub [295]
The total surface area of this triangular prism is 928cm
5 0
3 years ago
The mean number of students in a classroom at school a is 32.5 and there are 25 classrooms. The mean number of students at schoo
Airida [17]

We have given three school there, School A, School B and School C.

We are going to use following formula.

Total number of student in a school = Mean of number of student × Number of classrooms.

School A

Mean of students in School A = 32.5.

Number of classroom in School A = 25.

Number of students in School A = Mean × number of classrooms = 32.5×25 = 812.5


School B

Mean of students in School B = 29.6.

Number of classroom in School B = 12.

Number of students in School B = Mean × number of classrooms = 29.6×12 = 355.2


School C

Mean of students in School C = 15.3.

Number of classroom in School C = 10.

Number of students in School C = Mean × number of classrooms = 15.3×10 = 153.


Total sum of number of classes in all schools = 25+12+10 = 47.

Total sum of all students in all schools = 812.5 +355.2 +153 = 1320.7 .

The mean number of students per classroom for all the schools combined =

\frac{Total \ sum \ of \ all \ students \ in \ all \ schools}{Total \ sum \ of \ number \ of \ classes \ in \ all \ schools}

= \frac{1320.7}{47} = 28.1

The mean number of students per classroom for all the schools combined= 28.1

4 0
3 years ago
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