Thanks for the help. If this is the way to go, thank you, but I don't this is how it was taught in class. If anyone else has an alternative method, please let me know.tommyservo7 wrote:I tried. Looks like it requires a matrix.noname wrote: 2x+2z=2

5x+3y=4

3y-4z=4

The textbook says the solution should be (-4,8,5). How do you get this answer? I know it's something really simple, but I can't remember how to do it.

EDIT: This nifty little thing works excellently. It will teach you how to do it.

## Schoolwork Help Thread

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**2**of**2**- noname Offline
- ROAF
**Posts:**1852**Joined:**Tue May 06, 2008 4:14 pm

### Re: Schoolwork Help Thread

**Posted:** Sun Nov 28, 2010 8:33 pm

- AWA Offline
- MAN Chops
**Posts:**3939**Joined:**Sun Jan 04, 2009 1:35 pm**Location:**West Deptford, New Jersey, The United States of America, Earth, The Solar System, The Milky Way

### Re: Schoolwork Help Thread

**Posted:** Sun Nov 28, 2010 8:51 pm

noname wrote: 2x+2z=2

5x+3y=4

3y-4z=4

The textbook says the solution should be (-4,8,5). How do you get this answer? I know it's something really simple, but I can't remember how to do it.

It's a system of equations.2x+2z=2

5x+3y=4

3y-4z=4

The first equation (solved for z):

z=1-x

Substitute this statement into the third equation:

3y-4(1-x)=4

3y-4+4x=4

3y+4x=8

Now you have a simple system of two variables. Solve with the second equation (equation addition and subtraction):

3y+4x=8

3y+5x=4

-x=4

**x=-4**

Now substitute this x-value into the first equation:

2(-4)+2z=2

-8+2z=2

2z=10

**z=5**

You may choose to use either the x-value or the z-value to solve for y. Let us use the z-value for variety:

3y-4(5)=4

3y-20=4

3y=24

**y=8**

Putting our three results into an ordered triple:

(x,y,z)

**(-4,8,5)**

Check your answer. It should be correct.

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-Glenn Dawson

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- noname Offline
- ROAF
**Posts:**1852**Joined:**Tue May 06, 2008 4:14 pm

### Re: Schoolwork Help Thread

**Posted:** Sun Nov 28, 2010 9:06 pm

That's more like what we learned in class. Thanks to both of you for your help.AWA wrote:noname wrote: 2x+2z=2

5x+3y=4

3y-4z=4

The textbook says the solution should be (-4,8,5). How do you get this answer? I know it's something really simple, but I can't remember how to do it.It's a system of equations.2x+2z=2

5x+3y=4

3y-4z=4

The first equation (solved for z):

z=1-x

Substitute this statement into the third equation:

3y-4(1-x)=4

3y-4+4x=4

3y+4x=8

Now you have a simple system of two variables. Solve with the second equation (equation addition and subtraction):

3y+4x=8

3y+5x=4

-x=4

x=-4

Now substitute this x-value into the first equation:

2(-4)+2z=2

-8+2z=2

2z=10

z=5

You may choose to use either the x-value or the z-value to solve for y. Let us use the z-value for variety:

3y-4(5)=4

3y-20=4

3y=24

y=8

Putting our three results into an ordered triple:

(x,y,z)

(-4,8,5)

Check your answer. It should be correct.

- waffley Offline
- chops master
**Posts:**1266**Joined:**Tue May 11, 2010 2:59 am

### Re: Schoolwork Help Thread

**Posted:** Fri Feb 04, 2011 3:57 am

My homework will be due before anyone sees this, but here's a couple questions.

This one's multiple choice:

The average velocity of an object for a given motion points to the right. When is it possible for the instantaneous velocity of the object to point to the left? (Select all that apply.)

-at the beginning of the motion

-at some point in the middle of the motion

-at the end of the motion

I just 100% don't get this question and so I got it wrong.

Also this:

You drive in a straight line at 18.0 m/s for 12.0 miles, then at 30.0 m/s for another 12.0 miles.

(a) How does your average speed compare with 24.0 m/s?

speedav < 24.0 m/s

speedav > 24.0 m/s

speedav = 24.0 m/s

Explain.

I originally was like "oh I'm traveling different speeds over the same distance so I can just add those and take the average of that" which ended up being wrong. Later on in the question you calculate the average velocity and I realized that since you are traveling the same distance with different speeds, the time in which you travel those distances is different so you can't really take a shortcut there and you have to actually calculate out the distance traveled over time. Right? Does that make sense?

This one's multiple choice:

The average velocity of an object for a given motion points to the right. When is it possible for the instantaneous velocity of the object to point to the left? (Select all that apply.)

-at the beginning of the motion

-at some point in the middle of the motion

-at the end of the motion

I just 100% don't get this question and so I got it wrong.

Also this:

You drive in a straight line at 18.0 m/s for 12.0 miles, then at 30.0 m/s for another 12.0 miles.

(a) How does your average speed compare with 24.0 m/s?

speedav < 24.0 m/s

speedav > 24.0 m/s

speedav = 24.0 m/s

Explain.

I originally was like "oh I'm traveling different speeds over the same distance so I can just add those and take the average of that" which ended up being wrong. Later on in the question you calculate the average velocity and I realized that since you are traveling the same distance with different speeds, the time in which you travel those distances is different so you can't really take a shortcut there and you have to actually calculate out the distance traveled over time. Right? Does that make sense?

DCI open summer diddle band '11 '12

DCI world summer diddle band '13

WGI world winter band '12 '14

HS teachin' band '12-'15

DCI world summer diddle band '13

WGI world winter band '12 '14

HS teachin' band '12-'15

- ottomagne Offline
- MAN Chops
**Posts:**3114**Joined:**Mon Sep 14, 2009 1:24 am**Location:**Houston, TX

### Re: Schoolwork Help Thread

**Posted:** Fri Feb 04, 2011 11:30 am

I didn't quite understand the first one, but I do get the second one.waffley wrote:My homework will be due before anyone sees this, but here's a couple questions.

This one's multiple choice:

The average velocity of an object for a given motion points to the right. When is it possible for the instantaneous velocity of the object to point to the left? (Select all that apply.)

-at the beginning of the motion

-at some point in the middle of the motion

-at the end of the motion

I just 100% don't get this question and so I got it wrong.

Also this:

You drive in a straight line at 18.0 m/s for 12.0 miles, then at 30.0 m/s for another 12.0 miles.

(a) How does your average speed compare with 24.0 m/s?

speedav < 24.0 m/s

speedav > 24.0 m/s

speedav = 24.0 m/s

Explain.

I originally was like "oh I'm traveling different speeds over the same distance so I can just add those and take the average of that" which ended up being wrong. Later on in the question you calculate the average velocity and I realized that since you are traveling the same distance with different speeds, the time in which you travel those distances is different so you can't really take a shortcut there and you have to actually calculate out the distance traveled over time. Right? Does that make sense?

18 m/s = 12/t1, 12/18 s = t1

30 m/s = 12/t2, 12/30 s = t2

Average velocity = dd/dt, or 24/t1+t2

Because t1+t2 > 1 (12/18 is 2/3, 12/30 > 10/30 = 1/3), we know that the average velocity that you traveled will be < 24 m/s.

Drum bandin' since twenty o' seven.

'10 '13

'11'13

- waffley Offline
- chops master
**Posts:**1266**Joined:**Tue May 11, 2010 2:59 am

### Re: Schoolwork Help Thread

**Posted:** Sun Mar 20, 2011 10:18 pm

Posting this in here because everyone should know about this awesome guy who should probably be a tenured professor at some awesome university.

http://www.youtube.com/user/freelanceteach" onclick="window.open(this.href);return false;

http://www.youtube.com/user/freelanceteach" onclick="window.open(this.href);return false;

DCI open summer diddle band '11 '12

DCI world summer diddle band '13

WGI world winter band '12 '14

HS teachin' band '12-'15

DCI world summer diddle band '13

WGI world winter band '12 '14

HS teachin' band '12-'15

- petrolika Offline
- ramming notes
**Posts:**412**Joined:**Sun Sep 25, 2011 5:35 am**Location:**The Land of Oz.

### Re: Schoolwork Help Thread

**Posted:** Fri Sep 30, 2011 10:44 pm

Anyone here know anything about Agriculture, and knows how to do DSE (Dry Sheep Equivalent)??

Les femmes françaises ne sont pas poilu, ils ont juste beaucoup d'ombre autour de leurs aisselles, qui donne aux gens la perception de la pilosité français.

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