Answer:
5 of the bouquets are Daises.
Step-by-step explanation:
25 in all. 4/5 are Roses.
4/5 = Rose Bouquets 1/5 = Daisy Bouquets
4/5 = 20/25
1/5 = 5/25
So 5 are Daises
Answer:
54
Step-by-step explanation:
Answer:
x=-4
Step-by-step explanation:
These two angles have to equal 180 because they are opposite of each other. Therefore, the equation will be 129+x+x+59=180. You add all common terms. 2x+188=180. You want to isolate x so you have to subtract both sides by 188. Therefore, the equation will be 2x=-8. Divide both sides by 2 to isolate x. x=-8/2. This is equal to x=-4.
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(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>
Given:
The vertices of a triangle are R(3, 7), S(-5, -2), and T(3, -5).
To find:
The vertices of the triangle after a reflection over x = -3 and plot the triangle and its image on the graph.
Solution:
If a figure reflected across the line x=a, then
The triangle after a reflection over x = -3. So, the rule of reflection is
The vertices of triangle after reflection are
Similarly,
And,
Therefore, the vertices of triangle after reflection over x=-3 are R'(-9,7), S'(-1,-2) and T'(-3,-5).
