Fan and Medal!
The expression is shown below:

- anonymous

Fan and Medal!
The expression is shown below:

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- schrodinger

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- anonymous

Not sure if it is just me, but I am unable to see your expression : )

- anonymous

\[6a ^{2} - 5ab + 3b - 12\]

- anonymous

now do u see it

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## More answers

- anonymous

The question:
What is the-
a)Term
b)Constant
c)Coefficient

- anonymous

@jim_thompson5910

- anonymous

sorry to ask the question late i was helping another friend

- jim_thompson5910

the terms are separated by a + or a -

- jim_thompson5910

eg: `x+y-z+w` has four terms: x, y, z, w

- jim_thompson5910

the constant is any term that doesn't have a variable attached. It's a single fixed number

- anonymous

oh ok so there are 4 terms and 1 constant in the equation

- jim_thompson5910

yes

- anonymous

but what is coefficient

- jim_thompson5910

coefficients are the numbers in front of the variables

- jim_thompson5910

eg: in `7x`, the 7 is the coefficient

- anonymous

so there are 3 coefficients in the equation

- jim_thompson5910

yes

- anonymous

oh ok but its not done

- jim_thompson5910

I should have stated that subtraction is the same as adding a negative
so something like `1-2` is the same as `1 + (-2)`

- jim_thompson5910

therefore,
\[\Large 6a ^{2} - 5ab + 3b - 12\]
is the same as
\[\Large 6a ^{2} +(- 5ab) + 3b+( - 12)\]

- jim_thompson5910

in that second form, we see each term is separated by a +

- jim_thompson5910

terms are:
6a^2
-5ab
3b
-12

- anonymous

ok

- jim_thompson5910

I'm sure you can see what the constant is
for the coefficients, just list the numbers in front of each variable term

- anonymous

constant is 12

- jim_thompson5910

6,
NEGATIVE 5
3

- anonymous

y is it -5?

- jim_thompson5910

because the term is -5ab
the number in front is -5

- jim_thompson5910

go back to \[\Large 6a ^{2} +(- 5ab) + 3b+( - 12)\]

- anonymous

oh ok now i get it

- jim_thompson5910

the constant should be -12 for the same reason

- anonymous

constant=-12
coefficient=6, -5, 3
is that correct?

- jim_thompson5910

yep on all 4

- anonymous

thnx but its not done

- jim_thompson5910

what else is left?

- anonymous

they said to simplify 2 more expressions by combining like terms

- jim_thompson5910

which expressions?

- anonymous

ill show u

- anonymous

\[a) 5x-3-3x+6y+4\]

- jim_thompson5910

what are a pair of like terms that you see?

- anonymous

\[b) -8m ^{2} + 9m - 7m ^{2} + n ^{3}\]

- anonymous

the pair of like terms in the first one?

- jim_thompson5910

yeah do you see any like terms in the first one?

- anonymous

yes i see two pairs of like terms

- jim_thompson5910

ok one pair is what?

- anonymous

5x, 3x

- jim_thompson5910

more like 5x and -3x, but you have the right idea

- jim_thompson5910

they would combine to ???

- anonymous

u mean add?

- jim_thompson5910

yeah, 5x + (-3x) = ??

- anonymous

it's 2x

- jim_thompson5910

yep

- jim_thompson5910

you'll do the same for the other pair of like terms

- anonymous

what is the other pair

- jim_thompson5910

you tell me

- jim_thompson5910

look for any terms that have something in common

- jim_thompson5910

for the 5x and -3x, they had an x in common

- anonymous

is it -3, 4

- jim_thompson5910

yes

- anonymous

yay

- jim_thompson5910

they add to what?

- anonymous

is it 1?

- jim_thompson5910

yes

- anonymous

and then?

- jim_thompson5910

notice how there is only one y term, so there are no like terms for the y terms
all of this means `5x−3−3x+6y+4` simplifies to `2x+6y+1`

- anonymous

oh ok so 2x+6y+1 is the answer or should i evaluate the expression

- jim_thompson5910

if you knew the value of x and y, you could evaluate to get a single number

- jim_thompson5910

but they didn't provide values for x and y

- anonymous

So for now 2x+6y+1 is the answer, right?

- jim_thompson5910

yes

- anonymous

Ok thank u very much u dont have to explain the 2 expression because its almost the same as the first expression

- jim_thompson5910

yeah it's very similar so you shouldn't have too much trouble with it

- anonymous

thats it for today thank u very much dude ur the best

- jim_thompson5910

just keep in mind that terms with \(\Large m^2\) are NOT like terms with \(\Large m^3\)
they both have m, but different exponents makes them different and unlike terms

- jim_thompson5910

same with m and m^2

- anonymous

oh ok thanks for the tips i fanned and i gave u a medal for all three questions thank u very much i will ask u whenever i have a question from now on.

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