i need help with math. Algebra 9th grade.?

Question

tha problem says: Solve each formula for the indicated variable.

a. S = (n - 2) 180. Solve for n.

i feel guilty asking for help on my homework, but i just need to know how to do the first one. plz dont just give me the answer.

Answer 1

rewriting (n-2)180=S
n-2=S/180
n=(S/180)+2

Answer 2

n = S/180 + 2

Answer 3

What it seems like you have to do is isolate the n. That means put n to one side so that all the other numbers and variables are equal to it.

The first thing you can do is divide S by 180. This leaves you with S/180 = n-2

Then you have to get rid of the -2 on the right side so that only n is left behind so you add 2 to S/180.

The answer is up to you. Good luck.

Answer 4

I'm assuming that the equation reads
S equals (n minus two) times 180.

When you solve for any variable, your goal is to get it by itself on one side of the equation.

First, divide both sides by 180. (As long as you do the same thing to each side of an equation, it stays in balance, like a scale.)

S/180 = n-2

Then add 2 to both sides.

S/180 + 2 = n or, in proper form to answer the question,
n = S/180 + 2

Voila!

Answer 5

first you would distribute the 180 to each of the values in the parentheses and get:

S = 180n - 360

to solve for n, you would add the 360 to the S on the other side, and get

S + 360 = 180n

then divide by 180 on both sides and you'd get

(S/180) + 2 = n

good luck on the rest of your homework =)

Answer 6

S= (n - 2)180. Solve for n
S/180 = n-2
(S/180) +2 =n

Answer 7

Multiply 180 times n and 2

S=180n-360

Then add 360 to both sides.

S+360=180n

Then divide both sides by 180.

S+360
--------- = n
180

OR

S/180 + 2 = n

Whenever the question is "Solve for . . . " a variable, you should get that variable on one side of the equal sign all by itself.

Answer 8

okay you may want to discover 2 numbers that provide you the coefficient of the first and intensely last time period even as elevated or perhaps as further at the same time provide you the middle words coefficient.... there is something referred to as the field approach write the elements of the first coefficient in the first Q it is a million a million(x)^2 is your first time period..... a million -a million those are your elements a million and a million or -a million and -a million a million -a million try this for the perfect time period 6 a million 2 those are your elements a million and six and a pair of and three adverse 6 3 elements are includedtoo merely theres too many to jot down.. now u might want to discover the pair of elements from the first time period and the 2d time period that once elevated at the same time... and further provide you with the middle time period... thus 5x so that they might want to multiply and upload to 5 a million 3 that's trial and mistake a million 2 make a *field* such as your words and flow multiply so u get 3 x1 and a pair of x1 once you multiply them out you get 3 and a pair of wich upload to 5 your middle time period so that you recognize your suited.... then you definately merely position crackets round your elements (a million 3) (a million 2) those are literally your 2 elements.... the signal interior corresponds to the signal of the your elements. thus there both constructive so.. (x+3)(x+2)=0 set each and each to 0 x+3 = 0 or x = -3 x+2 = o x = -2

Answer 9

S = 180n - 360
s + 360 = 180n
(s+ 360) /180 = 180n/180
(s + 360) /180 = n

Answer 10

n= (S/180) + 2