All categories

4 years ago

Expand and simplify (2x+1) (x-2) (x+3)

Mathematics

2 answers:

nadezda [96]4 years ago

7 0

Answer:

2x^3 + 3x^2 - 11x - 6.

Step-by-step explanation:

(2x+1) (x-2) (x+3)

= (2x + 1)[x(x + 3)-2(x + 3)]

= (2x + 1)(x^2 + x - 6)

= 2x(x^2 + x - 6) + 1(x^2 + x - 6)

= 2x^3 + 2x^2 - 12x + x^2 + x - 6

= 2x^3 + 3x^2 - 11x - 6.

Mnenie [13.5K]4 years ago

6 0

Good morning,

Answer:

Step-by-step explanation:

(2x+1) (x-2) (x+3)

= (2x+1) [(x-2) (x+3)]

= (2x+1) [x²+3x-2x-6]

= (2x+1) [x²+x—6]

= 2x³+2x²-12x+x²+x-6

=2x³+3x²-11x-6.

You might be interested in

How do you think your culture would exist of there were no such thing as a social " norm?<br>

jasenka [17]

Answer:

Step-by-step explanation:

3 0

4 years ago

Allison earns $6,500 per month at her job as a principal. the chart below shows the percentages of her budget how much does Alli

shepuryov [24]

Like my answer and I’ll type the answer in the comments it could be a b c d e or I’ll type it out

3 0

3 years ago

A rectangular pyramid was sliced perpendicular to its base and through its vertex. What is the shape of the cross section shown

muminat

Answer:a triangle

Step-by-step explanation:trust me and if you get it right on the test give me a like

4 0

3 years ago

Read 2 more answers

Please help asap!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! will give brainliest!!!!!!

aksik [14]

Answer:

X is 40

Step-by-step explanation:

Look it up

3 0

3 years ago

Read 2 more answers

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

$P = {0.9545 - 0.6827}{2} = 0.1359$

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0

4 years ago