Answer:
9m, 12m, 15m
Step-by-step explanation:
Answer:
Therefore first number is 3 and second number is 18
Step-by-step explanation:
Sum of the two numbers is 21 and Second number is six times the first number
x+6x = 21
7x = 21
x = 21/7
x = 3
substitute x value in 6x to find second number,
= 6x
= 6*3
= 18
Answer:
10
Step-by-step explanation:
Given that:
Point X = (3, 2)
Point Y = (3, - 8)
Distance of point X to Y:
Distance = √(x2 - x1)^2 + (y2 - y1)^2
X1 = 3 ; y1 = 2 ; x2 = 3 ; y2 = - 8
Distance = √(3 - 3)^2 + (-8 - 2)
Distance = √(0)^2 + (-10)^2
Distance = √0 + 100
Distance = √100
Distance = 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>
