Hello there,
From Point A to Point B Peter traveled roughly 30 miles in 30 minutes.
From Point B to Point C Peter traveled roughly 130 miles in 2 hours and 30 minutes.
From Point C he didn't move in distance but the hours moved so c is out.
From Point C to Point D, Peter traveled roughly 180 miles in 3 hours.
Therefore D Peter had the fastest interval at D.
Given that the concentration has been modeled by the formula:
C(t)=50t/(t^2+25)
where:
t is time in hours.
The concentration after 5 hours will be given by:
t= 5 hours
plugging the value in the equation we get:
C(5)=(50(5))/(5^2+25)
simplifying the above we get:
C(5)=250/(50)=5 mg/dl
Answer: 5 mg/dl
Answer:
it could be a formula ex- l times w = area
Step-by-step explanation:
(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>
The semicircle is 25pi or 78.54
The rectangle is 120 (12*10)
The triangle is 24 (.5*12*4)
So the total is 222.54 (78.54+120+24)
