Question about Office Equipment & Supplies

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Posted on Jan 02, 2017

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SOURCE: the smallest of 3 consecutive odd integers is

Let x be the smallest integer, then x+2 is the middle and x+4 the largest one. Resolve the equation

x + 2 * ( x + 4 ) = 5 * ( x + 2 ) - 24

=> x = 11

The three numbers are therefore 11, 13, and 15. The fact they are odd is irrelevant for finding the answer.

Posted on Jan 17, 2011

SOURCE: Two consecutive even integers such that the

We'll call x the smaller integer and y the larger. So we know x+3y=22. We can rearrange this to say:

x=22-3y. Now we guess, I'll guess y=6, now we solve and get x=22-3(6)=4. 4 and 6 are two even consecutive integers! So the answer is 4 and 6.

Posted on Sep 15, 2011

SOURCE: find two consecutive integers if one-third of the

Of course it is 3 and 4! Are you testing our knowledge friend? ; )

Posted on Sep 19, 2011

SOURCE: Find two consecutive even integers such that twice

Let x be first (smaller) integer and y second (larger) integer.

- They are consecutive (and even) : y=x+2
- Twice the smaller exceeds the larger by 18: 2*x=y+18
- Substitute first equation into second: 2x=x+2+18
- 2x-x=2+18
- x=20
- y=x+2=22

These two numbers are 20 and 22.

Posted on Oct 09, 2011

The integers of the numbers on three raffle tickets are consecutive integers whose sum is 7,530 are 2509, 2510, and 2511.

Feb 24, 2015 | SoftMath Algebrator - Algebra Homework...

Let x be the smallest number.

Let x + 2 be the other number (consecutive even integer)

Now to translate the rest;)

three times the smaller 3(x)

19 more -19

sum of the two integers - (x) +( x+2)

Pulling it together,

3x -19 = x + x +2

collect like terms

3x - 19 = 2x + 2

Put all the constants on one side by adding 19 to both sides.

3x - 19 +19= 2x + 2 + 19

3x = 2x +21

Subtract 2x from both sides to have all the x's on one side.

3x - 2x = 2x +21 - 2x

x = 21

The other number is 21 + 2, or 23

Check:three times the smaller = 3 x 21 = 63

sum of the two integers = 21 + 23 or 44

is 63 at least 19 more than 44

Let x + 2 be the other number (consecutive even integer)

Now to translate the rest;)

three times the smaller 3(x)

19 more -19

sum of the two integers - (x) +( x+2)

Pulling it together,

3x -19 = x + x +2

collect like terms

3x - 19 = 2x + 2

Put all the constants on one side by adding 19 to both sides.

3x - 19 +19= 2x + 2 + 19

3x = 2x +21

Subtract 2x from both sides to have all the x's on one side.

3x - 2x = 2x +21 - 2x

x = 21

The other number is 21 + 2, or 23

Check:three times the smaller = 3 x 21 = 63

sum of the two integers = 21 + 23 or 44

is 63 at least 19 more than 44

Feb 19, 2015 | Office Equipment & Supplies

Let x be first (smaller) integer and y second (larger) integer.

These two numbers are 20 and 22.

- They are consecutive (and even) : y=x+2
- Twice the smaller exceeds the larger by 18: 2*x=y+18
- Substitute first equation into second: 2x=x+2+18
- 2x-x=2+18
- x=20
- y=x+2=22

These two numbers are 20 and 22.

Oct 03, 2011 | Office Equipment & Supplies

We'll call x the smaller integer and y the larger. So we know x+3y=22. We can rearrange this to say:

x=22-3y. Now we guess, I'll guess y=6, now we solve and get x=22-3(6)=4. 4 and 6 are two even consecutive integers! So the answer is 4 and 6.

x=22-3y. Now we guess, I'll guess y=6, now we solve and get x=22-3(6)=4. 4 and 6 are two even consecutive integers! So the answer is 4 and 6.

Sep 15, 2011 | Texas Instruments TI-84 Plus Calculator

Let x be the smallest integer, then x+2 is the middle and x+4 the largest one. Resolve the equation

x + 2 * ( x + 4 ) = 5 * ( x + 2 ) - 24

=> x = 11

The three numbers are therefore 11, 13, and 15. The fact they are odd is irrelevant for finding the answer.

x + 2 * ( x + 4 ) = 5 * ( x + 2 ) - 24

=> x = 11

The three numbers are therefore 11, 13, and 15. The fact they are odd is irrelevant for finding the answer.

Jan 17, 2011 | Texas Instruments TI-84 Plus Calculator

Let X be the smallest integer.

Let Y (equal to X+1) be the second integer.

Let Z (equal to X+2) be the third integer.

Then 2*X +0*Y- Z = 18

Substituting: 2*X - (X+2) = 18

Associative: (2*X - X) + 2 = 18

Substraction: X = 16

Solve for Y.

Solve for Z.

Note: check my work -- there *MAY* be an error in the above mathematics. :-)

Let Y (equal to X+1) be the second integer.

Let Z (equal to X+2) be the third integer.

Then 2*X +0*Y- Z = 18

Substituting: 2*X - (X+2) = 18

Associative: (2*X - X) + 2 = 18

Substraction: X = 16

Solve for Y.

Solve for Z.

Note: check my work -- there *MAY* be an error in the above mathematics. :-)

Oct 20, 2010 | Gateway Computers & Internet

X + ( X + 2 ) + ( X + 4 ) = 105

3X + 6 = 105

3X = 105 - 6

3X = 99

X = 99/3

X = 33

So 33 + 35 + 37 = 105

3X + 6 = 105

3X = 105 - 6

3X = 99

X = 99/3

X = 33

So 33 + 35 + 37 = 105

Oct 06, 2010 | American Standard Plumbing

try this website:

http://www.homeworkspot.com/

http://www.homeworkspot.com/

Sep 15, 2010 | Lufkin C213C 600" x 38" Long Consecutive...

The picture link didn't work for me... and seeing as you asked in August of 2008, I'm certain you don't need this answer, but here you go...

According to the TI-89 Guidebook:

This notation indicates an “arbitrary integer” that represents any integer. When an arbitrary integer occurs multiple times in the same session, each occurrence is numbered consecutively. After it reaches 255, arbitrary integer consecutive numbering restarts at @n0.

This notation indicates an “arbitrary constant” that represents any integer. When an arbitrary constant occurs multiple times in the same session, each occurrence is numbered consecutively. After it reaches 255, arbitrary integer consecutive numbering restarts at @0.

So it's the same concept, just a little different. It all depends if it is shown as @n1 or @1 where 1 can be any integer to 255

Hope this helps

According to the TI-89 Guidebook:

This notation indicates an “arbitrary integer” that represents any integer. When an arbitrary integer occurs multiple times in the same session, each occurrence is numbered consecutively. After it reaches 255, arbitrary integer consecutive numbering restarts at @n0.

This notation indicates an “arbitrary constant” that represents any integer. When an arbitrary constant occurs multiple times in the same session, each occurrence is numbered consecutively. After it reaches 255, arbitrary integer consecutive numbering restarts at @0.

So it's the same concept, just a little different. It all depends if it is shown as @n1 or @1 where 1 can be any integer to 255

Hope this helps

Aug 28, 2008 | Texas Instruments TI-89 Calculator

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