Answer:
2x + y + 5 = 0
Step-by-step explanation:
y − 1 = -2 (x + 3)
y − 1 = -2x − 6
2x + y + 5 = 0
Answer:
3/16
Step-by-step explanation:
3/8 * 4/8= 3/16
3/8= 3 pens out of 8 items
4/8= 4 crayons out of 8 items
Answer:
Step-by-step explanation:
We need to find the equation of the line perpendicular to the line 3x+2y=8 and passes through (-5,2).
The given line can be expressed as:
We can see the slope of this line is m1=-3/2.
The slopes of two perpendicular lines, say m1 and m2, meet the condition:
Solving for m2:
Now we know the slope of the new line, we use the slope-point form of the line:
Where m is the slope and (h,k) is the point. Using the provided point (-5,2):
(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>
Answer:
The Correct Answer is C.0.
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