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3 years ago

Factor the polynomial:81m+48mn

1 answer:

7 0

To factor this polynomial you will have to first find the GCF of 81 and 48 and its 3.

So then you put 3 out the parentheses. 3(----).

Then you have to see how many times 3 goes into 81m and 48m. It is 27m and 16mn.

So then this is what you get, 3(27m-16mn)

Hope it helps you!!!

You might be interested in

2(a³+b²+11)+1(3+a³+b+11)

Ann [662]

Answer:

$2(a^3+b^2+11)+1(3+a^3+b+11)=3a^3+2b^2+b+36$

Step-by-step explanation:

I assume that you need simplification of the given expression.

The given expression is:

$2(a^3+b^2+11)+1(3+a^3+b+11)$

Using distributive property and multiplying 2 inside the parenthesis and 1 inside the other parenthesis. This gives,

$2(a^3+b^2+11)=2\times a^3+2\times b^2+2\times 11\\2(a^3+b^2+11)=2a^3+2b^2+22\\\\1(3+a^3+b+11)=1\times 3+1\times a^3+1\times b+1\times 11\\1(3+a^3+b+11)=3+a^3+b+11=a^3+b+14$

Therefore, $2(a^3+b^2+11)+1(3+a^3+b+11)$ is equal to:

$2a^3+2b^2+22+a^3+b+14$

Now, combining like terms using the commutative property of addition, we get:

$=(2a^3+a^3)+2b^2+b+(22+14)\\=3a^3+2b^2+b+36$

Therefore, the simplified form is $3a^3+2b^2+b+36$

8 0

3 years ago

George finished 10 of 25 math problems.what percent of the problems did george finish

PIT_PIT [208]

Answer:

40%

Step-by-step explanation:

So 10/25=0.4

0.4=40%

8 0

2 years ago

Read 2 more answers

Which values are solutions to the inequality below? . . Check all that apply.. . x2 > 49. .

Marianna [84]

$x^2 \ \textgreater \ 49 \\ \\ 
x_{1,2} = \pm \sqrt{49} = \pm 7 \\ \\ 
\boxed{x \in (-\infty, ~-7)\bigcup (7,~ +\infty) }$

6 0

2 years ago

Read 2 more answers

Solve a=v/t for v<br> NEED HELP ASAP!!!

Sergeu [11.5K]

Answer:

$v = at$

Step-by-step explanation:

you need to cross multiple

3 0

2 years ago

if you stop at random in front of an animal exhibit,What is the probability that you will be looking at monkeys? I put the table

Mkey [24]

Answer:

1/10

step-by-step explanation:

----------------------------------

| animals | percent of animals |

----------------------------------

| goats | 16% |

----------------------------------

| camels | 4% |

----------------------------------

| giraffes | 6% |

----------------------------------

| llamas | 6% |

----------------------------------

| deer | 1% |

----------------------------------

| monkeys | 10% |

----------------------------------

| birds | 27% |

----------------------------------

| reptiles | 14% |

----------------------------------

| bears | 16% |

----------------------------------

( sorry if the table above isnt 100% right or straight ^ )

first you add up all the percents of how much animals are in this exhibit

16 + 4 + 6 + 6 + 1 + 10 + 27 + 14 + 16

= 100 %

then since we are looking for the probability of stopping in front of the monkeys, we take that 100 we got earlier and get the 10% of monkeys from our table

10/100

but since there isnt a 10/100 on the answers we look for something that equals to 10/100 and 1/10 does since you can just times both the 1 and 10 to equal the 10 and 100 you were looking for

1/10

making 1/10 your answer!

hope this helps! and if you need anymore help on this math question just ask me :)

4 0

2 years ago