Guide :

# Arrange and write the decimals in increasing order 14.367, -28.7784, 213.22, -361.238 ,5.2 ,-5.33

I need help I am in 6th grade

## Research, Knowledge and Information :

### Medford Public Schools Department of Mathematics

Medford Public Schools Department of Mathematics. ... Arrange and write the decimals in increasing order. ... −361.238, −28.7784, −5.33,5.2, 14.367, 213.22 b)

### Decimal Numbers - SlideShare

Compare and order decimal numbers. 5. ... How to read and write decimals or decimal numbers? ... Actual Difference/ Estimated Difference 7. 14.255 14.000 8. 28.267 ...

### math.uga.edu

367 6 4. 2 0. 3 0 1. 4 0 1. 5 0 1. 6 0 1. 7 0 1. 8 0 1. 9 0 1. 10 0 1. 11 0 1. 12 0 1. 13 0 1. 14 0 2. 15 0 ... 5. 22 5. 23 5. 24 5. 0. 1. 2. 3. 4. 5. 8 ...

### Run_Issue_14_1985_Feb by Zetmoon - issuu

... Run_Issue_14_1985_Feb, Author: ... P0KE1 98 , 4: POKE631 ,28:POKE63 2,21 1 :POKE633,157:POKE63 4,5. 204 ... per order. Money order. Write for a Complete free Catalog,

### Calaméo - Pre Calculo

... ᎏ or ᎏ 2 5 ᎏ Use point-slope form to write the ... 0 22. 5 22. 5 22. 5 ... TV Reading 20 8. 5 32 3. 0 42 1. 0 12 4. 0 5 14. 0 28 4. 5 33 7. 0 18 12. 0 ...

### Archived Problems - Project Euler

21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95 78 17 53 28 22 75 31 67 15 94 ... 1,3,5,15 21: 1,3,7,21 28: 1,2,4,7,14,28. ... not the order. {1, 2, 3, 4 ...

### OpenSourceChemistryCourseGrades9-12 | Ion | Acid

Feb 17, 2009 · OpenSourceChemistryCourseGrades9-12 ... 28) Write the correct formula for Barium Phosphate when Ba ... Eliminate 2 213 Type of reaction?:

### 31 CFR Ch. II (7-1-14 Edition) Fiscal Service, Treasury Title ...

Title 28 through Title 41 . ... call 202-741-6000 or write to the Director, ... The parts in these volumes are arranged in the following order: ...

### FRASER - St. Louis Fed - Discover Economic History

... .187 14,690,655 +i,oae,5»2 14.32l,0; ... Import«ofwheat.owt.l3,361.292 Imports of flour 2.727.642 ... 29% 3,501 22 July 31 34% Jan. • 28% ...

## Suggested Questions And Answer :

### write an equation that can be used to medel the data in table.

An equation that models the data is: y=(x^3)/2-2x^2+9x/2+1. To find the equation I used the fact that there are 4 data pairs and I can find 4 unknowns from 4 equations. I also know that if the ordered pair is (x,y), y increases with x, and y is defined for x in {0, 1, 2, 3}. Let y=ax^3+bx^2+cx+d. This equation has 4 unknown coefficients a to d. When x=0, y=1 so d=1. When x=1, y=4 so a+b+c+d=4 and a+b+c=3. c=3-(a+b). When x=2, y=6 so 8a+4b+2c+d=6 and 8a+4b+2(3-(a+b))=5. 6a+2b=-1, and b=-(6a+1)/2. Therefore c=3-(a-(6a+1)/2)=(4a+7)/2. When x=3, y=10 so 27a+9b+3c+1=10 and 27a-9(6a+1)/2+3(4a+7)/2=9. 54a-9-54a+12a+21=18 and 12a=6, so a=1/2. From a=1/2 we get b=-2 and c=9/2. If we substitute the data pairs we find that they satisfy the equation y=(x^3)/2-2x^2+9x/2+1. Therefore this equation models the data.

### How many different ways can Hector put 5 tulips in 2 vases?

Ordered pair (V1,V2) represents the number of tulips in the two vases. V1 and V2 identify the vases. The set of possibilities are: (5,0), (4,1). (3,2), (2,3), (1,4), (0,5) if the tulips are identical. If the tulips are distinct: A, B, C, D, E, the permutations are: (5,0): (ABCDE,0) [1] (4.1): (BCDE,A), (ACDE,B), (ABDE,C), (ABCE,D), (ABCD,E) [5] (3,2): (ABC,DE), (ABD,CE), (ABE,CD), (ACD,BE), (ACE,BD), (ADE,BC), (BCD,AE), (BCE,AD), (BDE,AC), (CDE,AB) [10] (2,3): (DE,ABC), (CE,ABD), (CD,ABE), (BE,ACD), (BD,ACE), (BC,ADE), (AE,BCD), (AD,BCE), (AC,BDE), (AB,CDE) [10] (1,4): (A,BCDE), (B,ACDE), (C,ABDE), (D,ABCE), (E,ABCD) [5] (0,5): (0,ABCDE) [1] [32 ways in total=1+5+10+10+5+1]  If the tulips in each vase can be placed in different orders, then there are more permutations: 2 ways of arranging 2 tulips; 6 ways of arranging 3; 24 ways for 4; 120 ways for 5. For example, for (BC,ADE) there are 2 ways to arrange BC, and 6 ways to arrange ADE. So (BC,ADE) spawn 2*6=12 arrangements. Since there are 10 (2,3) arrangements, each of these spawns 12, giving a total of 120 possible ways. For (1,4) each spawns 24, making 5*24=120 possible ways. For (0,5) again we have 120 ways. Since there are 6 rows above the grand total would be 6*120=720 ways.

### If 1=3,2=5,3=6,4=9 so how much word needed for equal of 5?

5=10 using the following logic. Switch to the binary system of counting: 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010 represent the numbers 1 to 10 decimal in binary. If we cross out those numbers with an odd number of ones we get: 11, 101, 110, 1001, 1010. When these are converted back to decimal we get: 3, 5, 6, 9, 10. These are the listed numbers in order: so 1=3, 2=5, 3=6, 4=9 and 5=10. Another way of solving the problem is to list the numbers in order that are the sum of an even number of powers of 2; or the sum of a pair of powers of 2. (1) 3=2+2^0; (2) 5=2^2+2^0; (3) 6=2^2+2^1; (4) 9=2^3+2^0; (5) 10=2^3+2^1; (6) 12=2^3+2^2, etc.

### (1, -20), (7, -20), (2, -40), (6, -40), (5, -50), (8, 10), (9, 46), (0, 7), (-1, 47), (-2, 89)

First put all the points (x,y) in order of the x value: -2 89 -1 47 0 7 1 -20 2 -40 5 -50 6 -40 7 -20 8 10 9 46   When x=0 y=7, so the constant term in the function is 7 (y intercept). There's a minimum (turning point) near x=5 because the values of y for x=2 and 6 (-40) are more positive than -50. The gradient (dy/dx) gets positively steeper as x increases after x=5, and negatively steeper between x=-2 and +2. The degree of the polynomial is even because the function is positive for larger magnitude positive and negative values of x. The zeroes are between x=0 and 1 and between x=7 and 8. The function can be split into two parts: (1) y for -2 Read More: ...

### how do you work out: 3+7=....(finite 5)

Not sure what you're asking here. Are you trying to sum 3 and 7 to a different base? In decimal the answer of course is 10. 10 means the base as a number to that base. If the base was 5, 10 would be the number 5 itself. If the base was 2, 10 would be the number 2. In base 5 we would count 0 1 2 3 4 10 11 12 13 14 20, corresponding to the numbers 0 to 10 in decimal. In base 5 we only use digits 0 to 4; in binary (base 2) we only have digits 0 and 1. In base 10 (decimal) we only have digits 0 to 9. In base 16 (hexadecimal) we have to invent symbols for digits we don't have: there's no single symbol after 9, so we use letters A to F to represent what we would write in decimal as 10 to 15. 10 in base 16 is the same as 16 in base 10. What's 16 in base 5? Well it's 3 times the base plus 1: 31. What is 16 in binary (base 2)? 10 stands for 2; 100 stands for 4 (2 squared); 1000 stands for 8 (2 cubed); 10000 stands for 16 (2^4). 3+7=20 in base 5 and the sum totally in base 5 would be 3+12=20.

### Eight horses are entered in a race...(Have more)

1(a) How many different orders are possible for completing the race? 8*7*6*5*4*3*2*1=8!=40,320 (b) In how many different ways can first, second, and third places be decided? (Assume there is no tie.) (8*7*6)=P(8,3)=336 or 6*C(8,3) where 6=3*2*1 the number of ways of arranging three items 2.)Telephone numbers consist of seven digits; the first digit cannot be 0 or 1. How many telephone numbers are possible? 8*10^6=8,000,000  3.)In how many ways can five people be seated in a row of five seats? 5*4*3*2*1=5!=120 4.)In how many ways can five different mathematics books be placed next to each other on a shelf? 5!=120 5.)In a family of four children, how many different boy-girl birth-order combinations are possible? (The birth orders BBBG and BBGB are different.) 16=2*2*2*2 from BBBB to GGGG 6.)Two cards are chosen in order from a deck. In how many ways can this be done if (a) the first card must be a spade and the second must be a heart?  13*13=169 (b) both cards must be spades? 13*12=156 7.)A company’s employee ID number system consists of one letter followed by three digits. How many different ID numbers are possible with this system? 26*10*10*10=26,000 Continued in comment...

### x+y+z=9,2x-3y+4z=13,2x-3y+4z=13,3x+4y+5z=40. solve

There are 4 equations but only 3 unknowns. This means one equation is unnecessary or there is inconsistency. Call the equations in order A, B, C and D. B and C are the same so we can remove one. Let's remove C. 3A+B: 3x+3y+3z+2x-3y+4z=27+13; 5x+7z=40. So 5x=40-7z. We have x in terms of z. D-3A: y+2z=40-27=13. So y=13-2z. We have y in terms of z. We can substitute for x and y in one equation (choose A) to leave z as the only variable: (40-7z)/5+13-2z+z=9. Multiply through by 5: 40-7z+65-10z+5z=45; -12z+60=0, so 12z=60 and z=5. We can now find x and y: 5x=40-7z=40-35=5, making x=1. y=13-2z=13-10=3. So the solution is x=1, y=3 and z=5. Substitute these values in the original equations to check them out.   Gauss-Jordan method: Write the equations in matrix format: [ 1 1 1 | 9 ] [ 2 -3 4 | 13 ] [ 3 4 5 | 40 ] R2→R2-2R1 [ 1 1 1 | 9 ] [ 0 -5 2 | -5] [ 3 4 5 | 40 ] R3→R3-3R1: [ 1 1 1 | 9 ] [ 0 -5 2 | -5 ] [ 0 1 2 | 13 ] R3→5R3+R2: [ 1 1 1 | 9 ] [ 0 -5 2 | -5 ] [ 0 0 12 | 60 ] R2→-R2+R3 then R2→(1/5)R2 and R3→(1/12)R3: [ 1 1 1 | 9 ] [ 0 1 2 | 13 ] [ 0 0 1 | 5 ] R2→R2-R3: [ 1 1 1 | 9 ] [ 0 1 1 | 8 ] [ 0 0 1 | 5 ] R1→R1-R2: [ 1 0 0 | 1 ] [ 0 1 1 | 8 ] [ 0 0 1 | 5 ] R2→R2-R3: [ 1 0 0 | 1 ] [ 0 1 0 | 3 ] [ 0 0 1 | 5 ] From this identity matrix x=1, y=3 and z=5.