Answer:
it's probably ours it's just a lucky guess
Step-by-step explanation:
The second term of the given sequence aₙ = -5(n² + 1n) when solved gives us; a₂ = -20
<h3>How to find the nth term of a sequence?</h3>
We are given the formula for a sequence as;
aₙ = -5(n² + 1n)
Where n is the position of the term.
Now, for the first term, we will have;
a₁ = -5(1² + 1(1))
a₁ = -10
The second term of the sequence is;
a₂ = -5(2² + 1(2))
a₂ = -20
Read more about Sequence at; brainly.com/question/6561461
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10 students of Elena's classmates like to make jewellery.
<u>Step-by-step explanation:</u>
Let total no.of classmates who liked the jewellery = x
Here, in the given problem, given she asked her classmates '20'. So, it clearly shows that total number of students/classmates = 20. Also, in that, 50% said like to make jewellery. Now find 'x' as below,
Hence, the total number of classmates who liked to make Jewellery = 10
(e) Each license has the formABcxyz;whereC6=A; Bandx; y; zare pair-wise distinct. There are 26-2=24 possibilities forcand 10;9 and 8 possibilitiesfor each digitx; yandz;respectively, so that there are 241098 dierentlicense plates satisfying the condition of the question.3:A combination lock requires three selections of numbers, each from 1 through39:Suppose that lock is constructed in such a way that no number can be usedtwice in a row, but the same number may occur both rst and third. How manydierent combinations are possible?Solution.We can choose a combination of the formabcwherea; b; carepair-wise distinct and we get 393837 = 54834 combinations or we can choosea combination of typeabawherea6=b:There are 3938 = 1482 combinations.As two types give two disjoint sets of combinations, by addition principle, thenumber of combinations is 54834 + 1482 = 56316:4:(a) How many integers from 1 to 100;000 contain the digit 6 exactly once?(b) How many integers from 1 to 100;000 contain the digit 6 at least once?(a) How many integers from 1 to 100;000 contain two or more occurrencesof the digit 6?Solutions.(a) We identify the integers from 1 through to 100;000 by astring of length 5:(100,000 is the only string of length 6 but it does not contain6:) Also not that the rst digit could be zero but all of the digit cannot be zeroat the same time. As 6 appear exactly once, one of the following cases hold:a= 6 andb; c; d; e6= 6 and so there are 194possibilities.b= 6 anda; c; d; e6= 6;there are 194possibilities. And so on.There are 5 such possibilities and hence there are 594= 32805 such integers.(b) LetU=f1;2;;100;000g:LetAUbe the integers that DO NOTcontain 6:Every number inShas the formabcdeor 100000;where each digitcan take any value in the setf0;1;2;3;4;5;7;8;9gbut all of the digits cannot bezero since 00000 is not allowed. SojAj= 9<span>5</span>
