find the quotient:
25c^4+20c^3
------------
5c

- anonymous

find the quotient:
25c^4+20c^3
------------
5c

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

when you divide with the same base you subtract the exponents 5c-c has an exponent of 1

- anonymous

25c^4/5c^1=5c^3 and 20c^3/5c^1=4c^2

- anonymous

so final answer is 5c^3+4c^2...do you understand?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

yea thanks

- anonymous

you're welcome...anything else?

- anonymous

yea is this right 63b^4-9b^3
---------
9b
= -7b^3+b^2

- anonymous

yes or 7b^3-b^2

- anonymous

good job!

- anonymous

thanks for ur help

- anonymous

u there?

- anonymous

-21w^4-21w^3
-------------
7w = -3w^3-3w^2 is this right?

- anonymous

i thnk i messed up it should be a + not - right?

- anonymous

sorry

- anonymous

you are correct!

- anonymous

u no much bout inequalities

- anonymous

yes

- anonymous

can u help me

- anonymous

I will try

- anonymous

y

- anonymous

y

- anonymous

n second one together

- anonymous

have to graph it too

- anonymous

do you know how to graph using y=mx+b with y-int and slope

- anonymous

n second one together

- anonymous

yea kind of confusing

- anonymous

ok second inequality is a horizontal line passing throug (0, 5) shading above the solid line

- anonymous

first one is a dotted line using y=mx+b the y-int(b=2) you start at (0,2) and use the slope(m-coefficient of x) which is 1/1 so from (0,2) rise 1(up) run 1(right) to get second point

- anonymous

shading should include (0,0) so to the right of the diagonal line

- anonymous

the solution is the area where the shading overlaps-top right area similar to less than shape : )

- anonymous

did you get lost?

- anonymous

yeah does it look like this kind of lol the direction
\
----------------
\

- anonymous

the diagonal lines is going up from left to right bc the slope is positive

- anonymous

like this lol
/
---------------------
/
/

- anonymous

so lower right is shaded right

- anonymous

##### 1 Attachment

- anonymous

top right---see pic but diagonal should be dotted

- anonymous

k thas what i have but had top line dotted too lol appreciate ur patience:)

- anonymous

##### 1 Attachment

- anonymous

dotted line for greater than or less than solid when you have equal sign with the inequality

- anonymous

love helping students in math so I have great patience : )

- anonymous

y>4
\[x \le2\]
up two
straight across on fourth line

- anonymous

2 is diagonal right dotted line

- anonymous

y>4 is dotted line through (0,4) shading above the line because when you substitute (0,0) into the inequality, 0>4 is not true so shading should not include test point (0,0)

- anonymous

2nd one is vertical solid line passing through (2,0) shade to the left because 0 is less than or equal to 2

- anonymous

solution area is in top left right angle

- anonymous

I think i got it right i would post pic but when other viewer clicks it opens blank dont no why.

- anonymous

what is your number i can send you a text with pic

- anonymous

nevermind I'll send to my email and post in here

- anonymous

but anyways thank you so much.... had to take a plato class fot algebra. i graduate may 26, this month so this will be my last open study. thanks again

- anonymous

CONGRATS!!! Do you need me to post pic

- anonymous

yea please

- anonymous

- anonymous

##### 1 Attachment

- anonymous

you can email me if you need any other math help else brishundra@yahoo.com

- anonymous

where is diagonal line at

- anonymous

no diagonal line is this one becuase y= is horizontal x=is vertical

- anonymous

so same is true for inequality graphs..graph them just like y= or x=

- anonymous

o yeah k got ur email thanks

- anonymous

welcome!

Looking for something else?

Not the answer you are looking for? Search for more explanations.