Solve the following formula for the specified variable.
C=1/4h(s+k) solve for s???

- anonymous

Solve the following formula for the specified variable.
C=1/4h(s+k) solve for s???

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- chestercat

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- anonymous

if this was a normal equation, and all those variables were numbers, how would you solve for s?
hint: try to get s by itself

- anonymous

It's been so long since i actually did this type of problem

- anonymous

i also have a question about the original question
is it C=(1/4)h(s+k)
or
C=1/(4h(s+k))

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## More answers

- anonymous

it's the first one that you put on there

- anonymous

c=1/4h(s+k)

- anonymous

the first and easiest thing to get rid of is the 1/4
if we want to re-write the equation, we could like this:
c=(h(s+k))/4
since we are dividing by 4 on the right, we can multiply both sides by 4 to get closer to having s by itself (by moving the 4 over)
we get:
4c=h(s+k)
now, to get rid of that h
notice its being multiplied by the quantity inside the parenthesis
to move it to the other side, we need to divide both sides by h
(4c)/h=s+k
finally, subtract k from both sides to get your answer

- anonymous

ok im confused

- anonymous

|dw:1351648217904:dw|

- anonymous

|dw:1351648246343:dw|

- anonymous

ok so when i solved that i got 4c/h-k

- anonymous

is that right ?

- anonymous

you should get 4c=h(s+k)

- anonymous

ok

- anonymous

That's what you got as the answer??

- anonymous

now, divide both sides by h

- anonymous

so the h on the one side would cancel and just leave me with s+k but i don't know what i would get on the other side

- anonymous

(4c)/h

- anonymous

so my answr is going to be 4c/h=s +k??

- anonymous

you have to get s by itself
subtract k from both sides

- anonymous

OMG!! I'm sooo lost

- anonymous

I thought that I just subracted something

- anonymous

it says solve for s
that means you have to get s by itself

- anonymous

think of it this way
if you have s+k and you want to get s by itself, wouldnt it be necessary to subtract k
s+k-k=s

- anonymous

yes

- anonymous

since you subtracted k from one side, you MUST subtract it from the other

- anonymous

ok im just confused on what im going to get i thought what i typed a minute ago was my answr

- anonymous

your answer is going to have s by itself on one side and a whole bunch of stuff on the other
after you subtract k from both sides, the final answer is s=((4c)/h)-k

- anonymous

4c/h-k?

- anonymous

I thought that is what i put a minute ago

- anonymous

it was not. you hadnt subtracted k yet

- anonymous

Ok well i got that one wrong because the C was not capital

- anonymous

b=1/2f(r+k)

- anonymous

for r

- anonymous

I've got this stuff due tommorrow and im still not doen after this

- anonymous

r u there?

- anonymous

yes. sorry

- anonymous

its ok. r u working it or something

- anonymous

i got distracted

- anonymous

for this one, you can follow the exact same steps as the first one. Multiply both sides by 2 to get rid of the 1/2
divide both sides by f
subtract k from both sides
what do you get?

- anonymous

one sec

- anonymous

i even wrote the wrong problem

- anonymous

its ok, it happens

- anonymous

b=1/2f(r+k)

- anonymous

And I'm solving for r

- anonymous

thats the exact same

- anonymous

i got 2c/f-k but that's not right

- anonymous

where did you get a c
theres no c in b=1/2f(r+k)

- anonymous

i have no idea

- anonymous

i got that one may i ask another onee

- anonymous

?

- anonymous

yes

- anonymous

|dw:1351649553338:dw|

- anonymous

that is my next one

- anonymous

hmmm
give me a sec

- anonymous

k

- anonymous

ok
start by simplifying sqrt((xy)^2)=xy
yes?

- anonymous

I guess i don't know lol

- anonymous

it is. trust me
anyway, on the top, you can simplify sqrt(x^5y^9) to sqrt(x^4y^8)*sqrt(xy)
sqrt(x^4y^8)=x^2*y^4
so
7x^2y^4*sqrt(xy)
----------------
3xy
one x and 1 y on the top and bottom cancel
answer:
(7xy^3*sqrt(xy))/3

- anonymous

|dw:1351649829505:dw|

- anonymous

can u draw that for me because it's hard for me to read and understand it

- anonymous

is that 3 suppose to be with the root?

- anonymous

or is it with the Y?

- anonymous

@etemplin

- anonymous

its with the y

- anonymous

were the original numbers with the root or were they outside?

- anonymous

because i put it with the y and it said it was wrong.

- anonymous

they were on the outside of the root but they were part of the root

- anonymous

were the original numbers (the 3 and 7) with the root or in front of it?

- anonymous

they were in front

- anonymous

|dw:1351650274050:dw|

- anonymous

that changes the answer then
i assumed they were not part of the root; i assumed they were multiplied by it
my mistake

- anonymous

like that

- anonymous

it's fine

- anonymous

\[\sqrt[7]{x ^{5}y ^{9}} or 7\sqrt{x ^{5}y ^{9}}\]

- anonymous

ok what is that

- anonymous

which is it?

- anonymous

is that the answer?

- anonymous

the first one

- anonymous

yea. when i solved it the first time i did it like the 2nd choice
the new answer is|dw:1351650440270:dw|

- anonymous

so my y is on the outside??

- anonymous

yes

- anonymous

you cant simplify the bottom at all, and the top can be just a little bit

- anonymous

what u mean

- anonymous

im just saying
the answer is the one i drew
the reason is given in the post after that

- anonymous

ok

- anonymous

that wasn't correct

- anonymous

is the 3 part of the root like the 7 is?

- anonymous

another answer is |dw:1351651467504:dw|

- anonymous

i apologize for any incorrect answers

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