Answer: 17
Step-by-step explanation:
3(5)+2
15+2
F(5)=17
Answer:
1. Identify the problem: Packaging boxes use too much material and create waste
2. What are the equations for the volume and surface area of a cube and rectangular prism?
Volume of a cube: Vcube = L x L x L = L3
Surface area of a cube: SAcube = 6 x (L x L) = 6L2
Volume of a rectangular prism: VRP = L x W x H = LWH
Surface area of a rectangular prism: SARP = 2 x (L x W) + 2 x (L x H) + 2 x (W x H) = 2(LW + LH + WH)
3. What is the difference in surface area of the packages below? (Note that they have the same volume.)
SAcube = 6L2
= 6 (20 cm)2
= 2,400 cm2
SARP = 2(LW + LH + WH) = 2 (20cm x 10cm + 20cm x 40cm + 10cm x 40cm) = 2,800 cm2
SARP – SAcube = 2,800 cm2
– 2,400cm2
= 400 cm
Step-by-step explanation:
everything in bold is the answer
Can I please get the Brainlist
Answer:
x = 1/6
Step-by-step explanation:
3x + 4 = 9x + 3
3x + 1 = 9x
1 = 6x
1/6 = x
Suppose:
C=cost
c=cost per kilo-watt hour
w=number of watts
h=number of hours
thus
C=kwhc
where:
k=constant of proportionality:
k=C/(whc)
but
C=15 when w=6000, h=10/60=1/6 hours, c=7.5 cents per kw/h
thus
k=15/(6000*1/6*7.5)
k=0.002
thus:
C=0.002(whc)
hence:
when
h=70*2=140=140/6=70/3 hours the cost will be:
C=0.002(6000*70/3*7.5)
C=2100 cents
C=$21
(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>
