math homework help?
here is the problems can you also explain
1.there are two sizes of tables in a banquet hall. one size seats exactly 5 people and the other size seats exactly 8 people. last night, 66 people were seated at fewer than 10 tablas with no empty seats. how many tables of each size were there?
2.since prehistorical times, baskets have been made by people of all countries. American Indians make baskets for storing and carrying food. find all the possible row and column combinations for a rectangular display of 64 baskets made by the Promo Indians of California.
3.when a marching band marches in 4 or 5 columns, there is one marcher left over. what is the least number of band members possible?
4.if two kinds of bricks have heights 12cm and 20 cm respectively, what is the least number of rows of each kind that will have equal heights?
5.a jewelry maker made 4 silver rings and 6 gold rings.she wants to package them in groups of equal size of the same type.how many should he put in each package?
6.a jewelery maker made 12 pairs of emerald earrings 15 pairs of ruby earrings, and 18 pairs of diamond earrings. he wants to package them in groups of equal size of the same type. how many should he put in each package?
sorry i for got a problem
thanks
<3
for number 1 can you please wirte it in sentecne form
thanks
a lot
<3
1.) You set "x" for the number of 5-person tables and "y" for the number of 8-person tables. First equation: you know that "x + y = 13." Second equation: you know that "5x + 8y = 66." From the first equation you know that "x = 13 – y." You plug that into the second equation to get 5(13 – y) + 8y = 66. Solve for "y" and then you can solve for "x" and you’re good.
2.) First off, you probably already know that in a rectangular setup, "rows x columns = total." So your possible row or column numbers are all factors of 64. Factors of 64 are "1,2,4,8,16,32,64." 1 always goes with 64, 2 always goes with 32, 4 always goes with 16, and 8 always goes with 8. So you have 4 possible pairs. But for each pair (except 8 and 
one of the numbers can be either a row or column number. So here are the possible combinations:
rows: 1 columns: 64
columns: 1 rows: 64
rows: 2 columns: 32
columns: 2 rows: 32
rows: 4 columns: 16
columns: 4 rows: 16
columns: 8 rows: 8
So you should end up with 7 different combinations.
3.) You want the number to be as low as possible. This number has to be a multiple of 4 or a multiple of 5 with an additional +1 (cause you have one left over). Since 4 is smaller we should work with it (since we want a small number). If the number was 4 x 1, then we would add 1 to it and get 5. But 5 divided by 5 has nothing left over. So let’s try 4 x 2 = 8. 8 + 1 = 9. but 9 divided by 5 gives 4 left over. lets try 4 x 3 = 12. 12 + 1 = 13. 13 divided by 5 gives 3 left over. still not good. 4 x 4 = 16. 16 + 1 = 17. 17 divided by 5 gives 2 left over. 4 x 5 = 20. 20 + 1 = 21. 21 divided by 5…. gives 1 left over! So the number should be 21.
21 divided by 4 gives us 5 with one left over. 21 divided by 5 gives us 4 with one left over.
4.) This is kind of like number 3. We’re trying to get a small number that 12 and 20 will both go into. So lets use the smaller number 12. We’re not adding one because we’re not dealing with left overs. We want a multiple of 12 that will equal a multiple of 20. Lets start from the bottom. 12 x 1 = 12. 12 / 20 doesn’t really work, because 20 can’t go into 12 nicely. 12 x 2 = 24. 24 / 20 gives us left overs. 12 x 3 = 36. 36 / 20 gives us left overs. 12 x 4 = 48. 48 / 20 gives us left overs. 12 x 5 = 60. Does 20 go into 60? Yup. So we started with a small number (12) and 60 was as small as we can get and still include 20 as a factor that goes into it.
So, how many rows of 12 cm blocks go into 60 cm? we get 5.
How many rows of 20cm blocks go into 60 cm? we get 3.
5.) So we want packages of equal size. We know that we can’t have more than 4 packages. There wouldn’t be enough silver rings to go around. We can’t cut the rings so we know we’re gonna have 4 packages or less. Lets say we want 4 packages. We can’t do that. 6 can’t be divided into 4 cleanly. We get left overs (which means we had to cut rings… not possible). So lets try 3 packages. We can’t do that because that would mean cutting the silver rings ( 4 / 3 = left overs). So lets try 2 packages. That works! We can put 2 silver rings and 3 gold rings in each package. Not very many packages but it’s the only way to work it out. You COULD do one package with 4 silver rings and 6 gold rings, but the question asks for group"s" of equal size, meaning more than one.
(1)
Let s be number of small tables
Let b be number of big tables
5s + 8b = 66
And b needs to be one more than a multiple of 5 (since 5 people can sit at the small table)
b = 2 or 7
If b = 2, then s has to be 10. That would make 12 tables which is too many.
So b = 7, which means s = 2.
(3) So when you divide the number of marchers by 4 and 5, you have a remainder of 1 in both cases. Whats the smallest number you can find that at?
21
21 / 4 = 5 remainder 1
21 / 5 = 4 remainder 1
(5) So she should make packages of 24. This way she could package 6 sets of 4 silver rings, and 4 sets of 6 gold rings.
References :
1. (7 x
= 56 + (2 x 5) = 10 10 + 56 = 66
References :
ok for number 1 there were 7 tables of 8 and 2 tables of 5. If you multiply 8*7 (8 per table and 7 tables) you get 56 subtract it from 66 (total people) you get 10 which is 2*5 (2 tables with 5 people at each)
sorry i can’t help with the others
References :
1.
Let x = number of 5 seat tables
Let y = number of 8 seat tables
x + y < 10
5x + 8y = 66
We know there has to be 9 tables as even 8 of the larger (8 seat) tables would only seat 64, so:
x + y = 9
5x + 8y = 66
Solve for x in the first equation:
x = 9-y
Now substitute this into the second:
5x + 8y = 66
5(9-y) + 8y = 66
45 – 5y + 8y = 66
45 + 3y = 66
3y = 21
y =7
x = 9-y = 9-7 = 2
2.
This would be a permutation of a*b = 64:
a = 1, b = 64
a = 2, b = 32
a = 4, b = 16
a = 8, b = 8
a = 16, b = 4
a = 32, b = 2
a = 64, b = 1
3.
4*5 + 1 = 21
4.
12a = 20b
a = 5, b = 3
12*5 = 20*3
60 = 60
5.
4 can factor into 2 * 2
6 can factor into 2 * 3
The only common factor is 2, so she should put 2 silver and 3 gold rings into each package and have only 2 packages
References :
1.) You set "x" for the number of 5-person tables and "y" for the number of 8-person tables. First equation: you know that "x + y = 13." Second equation: you know that "5x + 8y = 66." From the first equation you know that "x = 13 – y." You plug that into the second equation to get 5(13 – y) + 8y = 66. Solve for "y" and then you can solve for "x" and you’re good.
2.) First off, you probably already know that in a rectangular setup, "rows x columns = total." So your possible row or column numbers are all factors of 64. Factors of 64 are "1,2,4,8,16,32,64." 1 always goes with 64, 2 always goes with 32, 4 always goes with 16, and 8 always goes with 8. So you have 4 possible pairs. But for each pair (except 8 and
one of the numbers can be either a row or column number. So here are the possible combinations:
rows: 1 columns: 64
columns: 1 rows: 64
rows: 2 columns: 32
columns: 2 rows: 32
rows: 4 columns: 16
columns: 4 rows: 16
columns: 8 rows: 8
So you should end up with 7 different combinations.
3.) You want the number to be as low as possible. This number has to be a multiple of 4 or a multiple of 5 with an additional +1 (cause you have one left over). Since 4 is smaller we should work with it (since we want a small number). If the number was 4 x 1, then we would add 1 to it and get 5. But 5 divided by 5 has nothing left over. So let’s try 4 x 2 = 8. 8 + 1 = 9. but 9 divided by 5 gives 4 left over. lets try 4 x 3 = 12. 12 + 1 = 13. 13 divided by 5 gives 3 left over. still not good. 4 x 4 = 16. 16 + 1 = 17. 17 divided by 5 gives 2 left over. 4 x 5 = 20. 20 + 1 = 21. 21 divided by 5…. gives 1 left over! So the number should be 21.
21 divided by 4 gives us 5 with one left over. 21 divided by 5 gives us 4 with one left over.
4.) This is kind of like number 3. We’re trying to get a small number that 12 and 20 will both go into. So lets use the smaller number 12. We’re not adding one because we’re not dealing with left overs. We want a multiple of 12 that will equal a multiple of 20. Lets start from the bottom. 12 x 1 = 12. 12 / 20 doesn’t really work, because 20 can’t go into 12 nicely. 12 x 2 = 24. 24 / 20 gives us left overs. 12 x 3 = 36. 36 / 20 gives us left overs. 12 x 4 = 48. 48 / 20 gives us left overs. 12 x 5 = 60. Does 20 go into 60? Yup. So we started with a small number (12) and 60 was as small as we can get and still include 20 as a factor that goes into it.
So, how many rows of 12 cm blocks go into 60 cm? we get 5.
How many rows of 20cm blocks go into 60 cm? we get 3.
5.) So we want packages of equal size. We know that we can’t have more than 4 packages. There wouldn’t be enough silver rings to go around. We can’t cut the rings so we know we’re gonna have 4 packages or less. Lets say we want 4 packages. We can’t do that. 6 can’t be divided into 4 cleanly. We get left overs (which means we had to cut rings… not possible). So lets try 3 packages. We can’t do that because that would mean cutting the silver rings ( 4 / 3 = left overs). So lets try 2 packages. That works! We can put 2 silver rings and 3 gold rings in each package. Not very many packages but it’s the only way to work it out. You COULD do one package with 4 silver rings and 6 gold rings, but the question asks for group"s" of equal size, meaning more than one.
References :