Oh, math genius'... I am back for help

letfuller

<font color=red>The scheming queen for disney trip
Joined
Jan 12, 2003
Messages
620
I know that I should know and I really am just checking my answer to the math genius'


Solve: 6x+4(3x-2)=6x+2



Also:

The length of a rectangle is 1 in. more than twice its width. If the perimeter of the rectangle is 50 in., find the width of the rectangle. Solve using an algebraic equation.



Oh.... THANK YOU
 
I know that I should know and I really am just checking my answer to the math genius'


Solve: 6x+4(3x-2)=6x+2



Also:

The length of a rectangle is 1 in. more than twice its width. If the perimeter of the rectangle is 50 in., find the width of the rectangle. Solve using an algebraic equation.



Oh.... THANK YOU

Are you familiar with the distibutive principal? You need it for the first one!
 
For the first one I got x = 5/6.

The second one:

L = 2W + 1
2L + 2W = 50

2(2W + 1) + 2W = 50

4W + 2 + 2W = 50

6W = 48

W = 8

L = 2(8) + 1

L = 17

Let me know if you don't understand something.

Kimya
 

6x+4(3x-2)=6x+2

6x+ 12x - 8 = 6x+2

18x -8 = 6x + 2

12x = 10

x = 5/6

(same answer as above, just showing the steps :) )
 
Ooohh! math people! can you guys help me with my test review?

K^2+5k+6 888 88 k^2 + 9k
----------- *** * ----------
k^2+12k+27 888888 k^2-2k-8

I'm supposed to multiply and simplify but I missed that class and I'm so lost!!

the numbers in white are just so it would look right lol! Thanks for any help you guys can give me!
 
isn't it Thanksgiving break????


Alright...

6x+4(3x-2)=6x + 2
6x +12x-8=6x + 2
18x - 8 = 6x + 2
12x - 8 = 2
12x = 10
x=10/12
x = 0.83
(I showed all of the steps individually...but you could simplify)

Next one:

The length of a rectangle is 1 in. more than twice its width. If the perimeter of the rectangle is 50 in., find the width of the rectangle. Solve using an algebraic equation.

width = x
length = 2x + 1

perimeter = 2x + 2(2x + 1) = 50
2x + 4x + 2 = 50
6x + 2 = 50
6x = 48
x = 8 inches
lenght then would be 17 inches...let's check it:
8+8+17+17= 50
 
Ooohh! math people! can you guys help me with my test review?

K^2+5k+6 888 88 k^2 + 9k
----------- *** * ----------
k^2+12k+27 888888 k^2-2k-8

I'm supposed to multiply and simplify but I missed that class and I'm so lost!!

the numbers in white are just so it would look right lol! Thanks for any help you guys can give me!

(k+2)(k+3) k(k+9)
________ x ______
(k+3)(k+9) (k-4)(k+2)

cancel out like terms on top and bottom (k+3) (k+2) (k+9)

leaves you with k/(k-4)

Kimya
 
I was right on track! Thank you oh sage mathmaticians. I should just stick to my job teaching English!!!
 
(k+2)(k+3) k(k+9)
________ x ______
(k+3)(k+9) (k-4)(k+2)

cancel out like terms on top and bottom (k+3) (k+2) (k+9)

leaves you with k/(k-4)

Kimya

:banana: thank you so much!!! you really are a math Genius:thumbsup2
 
:banana: thank you so much!!! you really are a math Genius:thumbsup2

It was always my best subject! I finished up through MultiVariable Calculus with A's in all of my Calc classes.

Now History on the other hand.....I'd probably end up telling you that the War of 1812 happened in 1492. :laughing:

Kimya
 
It was always my best subject! I finished up through MultiVariable Calculus with A's in all of my Calc classes.

Now History on the other hand.....I'd probably end up telling you that the War of 1812 happened in 1492. :laughing:

Kimya

:rotfl: I LOOOOVE history. i think i would do ok in math but I have a HORRIBLE prof. he expects everyone to understand what to do just by him reading the book at us:headache: and maybe showing us on the board once. But the semesters almost over and I just have one more test and the final to get through
 
:rotfl: I LOOOOVE history. i think i would do ok in math but I have a HORRIBLE prof. he expects everyone to understand what to do just by him reading the book at us:headache: and maybe showing us on the board once. But the semesters almost over and I just have one more test and the final to get through

Would it make you sick if I told you I can almost always read it in the math book and understand it just fine? :teeth:

Kimya
 
One more story problem that I just can't figure out. I never could in high school, undergrad.....



Find two consecutive integers such that the sum of 2 times the first integer and 7 times the second integer is 88. Solve using an algebraic equation.
 
One more story problem that I just can't figure out. I never could in high school, undergrad.....



Find two consecutive integers such that the sum of 2 times the first integer and 7 times the second integer is 88. Solve using an algebraic equation.

2X + 7(X+1)=88
9X +7=88
9X=81
X=9

answer: 9, 10
 
One more story problem that I just can't figure out. I never could in high school, undergrad.....



Find two consecutive integers such that the sum of 2 times the first integer and 7 times the second integer is 88. Solve using an algebraic equation.

Your integers are X and (X + 1) since they are consecutive.

2(x) + 7(x+1) = 88
2x + 7x + 7 = 88
9x = 81
x=9
x+1 = 10

try it and see

2(9) + 7(10) = ?
18 + 70 = 88

Kimya

ETA: Punkin beat me!
 
Oh come on man! I thought it was Thanksgiving break!

And besides, you all might as well be speaking Greek, that's how horrible I was at math. the first problem I probably could have managed if I had paper and pencil.

The rest ........:confused3 .........the math moron strikes again!
 
I am just thankful for all of your help. I really should just stick to reading novels and writing essays.
 
The first one is probably the only one I could do. Pretty bad at math here.
 

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