Answer:
B=![\left[\begin{array}{ccc}0&0\\0&1\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%5C%5C0%261%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Let's do the multiplication AB.
If A=![\left[\begin{array}{ccc}1&0\\0&0\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%260%5C%5C%5Cend%7Barray%7D%5Cright%5D)
then the first row of A is= (1 0) by the first column of B= (0 0) is equal to zero.
the first row of A is= (1 0) by the second column of B= (0 1) is equal to zero too because 1.0+0.1=0.
the second row of A is= (0 0) by any colum of B is equal to zero too.
So we have found an example that works!
Answer:
40
Step-by-step explanation:
30/75 = .4
Answer:
The graph on the far right
Step-by-step explanation:
The negative in front of the absolute value indicates that the parent function |x| has been reflected over the x-axis and points downwards. The negative 2 being added to this portion shows that there has been a vertical shift of two units downwards as well. Therefore, the graph on the far right is the correct graph.
740 + 850 x 100=? And there!!
(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>
