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[Constellation.]

TYPE OF QUESTION: Home Arts
YOUR QUESTION:Why is it important to keep the cookies on the baking sheet for at least a minute after taking them out of the oven?
Why is it important to cool the cookies completely before packaging?

We have a new teacher and she didn't even tell us any of it.. and i couldn't find it on any websites. I was getting really frustrated about it when i remembered this topic.

I know some cookies will go soggy if you seal them immediately. As for leaving the cookies on the baking sheet, I think it's mainly just so you don't cook your fingers along with the cookies... I'd like a second opinion on that though.

I bake and I know if you take them off the sheet they fall right apart.
I learned from experience they bake a bit outside the oven. I let them bake fully in the oven and they came out "well done".
But I'm no expert, may need a second opinion if not too late. ❤️

[Waki]

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[dakotapaws]

TYPE OF QUESTION:(CP Algebra 2)
YOUR QUESTION:(I actually have three different question on the same subject:
1. 3x+y+z=14
X+2y-3z=-9
5x-y+5z=30
(1,5,6)
2. 2x-y+2z=-7
-x+2y-4z=5
X+4y-6z=-1
(-3,-1,-1)
3. 3x-y+2z=4
6x-2y+4z=-8
2x-y+3z=10
(No solution)
It says to Solve the system using the elimination method. It's chapter 3.4 in the book ;3; If you have any questions about this you can pm me if you'd like ^-^ I do know the answers but I don't know how to get them, that's what I'm having trouble with.)

[TheFae]

TYPE OF QUESTION:(CP Algebra 2)
YOUR QUESTION:(I actually have three different question on the same subject:
1. 3x+y+z=14
X+2y-3z=-9
5x-y+5z=30
(1,5,6)
2. 2x-y+2z=-7
-x+2y-4z=5
X+4y-6z=-1
(-3,-1,-1)
3. 3x-y+2z=4
6x-2y+4z=-8
2x-y+3z=10
(No solution)
It says to Solve the system using the elimination method. It's chapter 3.4 in the book ;3; If you have any questions about this you can pm me if you'd like ^-^ I do know the answers but I don't know how to get them, that's what I'm having trouble with.)

These are actually really fun (but then again I love math!)!

so to start off (we are only working with problem 1!), you want to use one equation (say the first equation). You solve for one variable. Let's solve for 'z' because it's easier. so z = 14 - 3x - y.
In the next two problems, you then plug in for your z.
so:
x + 2y - 3(14-3x-y)=-9
5x - y + 5(14-3x-y)=30

From there you distribute and reduce down! (if you want exact steps, feel free to PM me)
So your new equations are:

10x + 5y = 33
-10x - 6y = -40

From there, solve for your variables.

I actually get x= -1/5 y=7 and z= 38/5
I even checked on my calculator, so you might have mistyped something! But this is the overall gist!
It's a lot of work, but I find it to be really fun!

I hope I helped!

[aequilibrium]

@TheFae - actually, that's not the elimination method, it's the substitution method :s

TYPE OF QUESTION:(CP Algebra 2)
YOUR QUESTION:(I actually have three different question on the same subject:
1. 3x+y+z=14
X+2y-3z=-9
5x-y+5z=30
(1,5,6)
2. 2x-y+2z=-7
-x+2y-4z=5
X+4y-6z=-1
(-3,-1,-1)
3. 3x-y+2z=4
6x-2y+4z=-8
2x-y+3z=10
(No solution)
It says to Solve the system using the elimination method. It's chapter 3.4 in the book ;3; If you have any questions about this you can pm me if you'd like ^-^ I do know the answers but I don't know how to get them, that's what I'm having trouble with.)

I'll be doing the second one:
2x-y+2z=-7
-x+2y-4z=5
x+4y-6z=-1

With this one you can cut the x's on the second and third equations:
-x+2y-4z=5
x+4y-6z=-1
6y - 10z = 4- simplify
3y - 5z = 2

You can replace one of the two equations (I'll just go with the last one) and then multiply the second one by 2 (or divide the first one by 2 but it's easier to multiply) to cut the x's:
2x-y+2z=-7........ 2x - y + 2z = -7
-x+2y-4z=5........-2x + 4y - 8z = 10
3y - 5z = 2........ ------

2x - y + 2z = -7
-2x + 4y - 8z = 10
3y - 6z = 3 - you can simplify it
y - 2z = 1

Replace it in the system (with the first or the second equation) and repeat the process until you get the answer:
2x - y + 2z = -7..... ------
y - 2z = 1............ -3y + 6z = -3
3y - 5z = 2........... 3y - 5z = 2
........................ z = -1

And then keep soliving it for the x and then for the y by replacing the z in the second equation and then the x you get there and the z in the first eQuat ion and it's done. I don't have more time right now, but if you want me to I can finish this one and the others later (in + or - 3-4 hours). I hope this helped you! c:

[fika.]

TYPE OF QUESTION: Math
YOUR QUESTION:make x the subject of the formula

y = x / a

i probably seem so stupid for asking this, but fractions are my weakest point in math and the numbers just fly around my head ;v;

[aequilibrium]

TYPE OF QUESTION: Math
YOUR QUESTION:make x the subject of the formula

y = x / a

i probably seem so stupid for asking this, but fractions are my weakest point in math and the numbers just fly around my head ;v;

Well, all you have to do is isolate the x. Since the x is divided by a, you need to switch a from one side to the other, so y multiplies by a.

y.a = x _ or _ x = y.a (same thing but with x on the left)

another way to explain this is that you have to get rid of everything that is with the x. the x is divided by a and the only way to get rid of it is multiply by a because x / a . a = x. Since this is an equation, what you do on one side must also be done on the other side so both sides are multiplied by a and you get y.a = x.

I thought explaining it would be nice for you to understand why and how, i'm sorry that i'm not that good at explaining things cx
I hope this helped!

[Waki]

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[Waki]

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[Waki]

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