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2 years ago

Factor completely, the expression: 2x^3-2x^2-12x

1 answer:

muminat2 years ago

8 0

$2x^3-2x^2-12x=2x(x^2-x-6)=(*)\\\\------------------------------\\x^2-x-6\\\\a=1;\ b=-1;\ c=-6\\\\\Delta=(-1)^2-4\cdot1\cdot(-6)=1+24=25\\\\\sqrt\Delta=\sqrt{25}=5\\\\x_1=\frac{1-5}{2\cdot1}=\frac{-4}{2}=-2;\ x_2=\frac{1+5}{2\cdot1}=\frac{6}{2}=3\\\\x^2-x-6=(x+2)(x-3)\\\\------------------------------\\\\(*)=2x(x+2)(x-3)$

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2 years ago

Convert 20 miles in km

Tamiku [17]

Answer:

32.1869

Step-by-step explanation:

Formula :

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1 year ago

a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was

Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

Let $\bar X$ = sample average time spent studying

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = population mean hours spent studying = 25 hours

$\sigma$ = standard deviation = 15 hours

n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < $\bar X$ < 30 hours)

P(29 hours < $\bar X$ < 30 hours) = P( $\bar X$ < 30 hours) - P( $\bar X$ $\leq$ 29 hours)

P( $\bar X$ < 30 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ < $\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z < 2) = 0.97725

P( $\bar X$ $\leq$ 29 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 29-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z $\leq$ 1.60) = 0.94520

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.

Therefore, P(29 hours < $\bar X$ < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

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2 years ago

It can be multiple answers and explain how to figure this out

Leno4ka [110]

None of the two answer choices are correct, answer is none of the above

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7 0

2 years ago

Right △ABC has coordinates A(-7, 3), B(-7, 10), and C(-1, 3). The triangle is reflected over the x-axis and then reflected again

aalyn [17]

X-axis = A (-7,-3) B (-7,-10) C (-1,-3) first they changed to the x-axis which is this. Just so you know on the x-axis the x-axis stay and the y-axis changed which is this.
When they reflect again on the y-axis the verter of A' is
A' (7, 3) = answer
On the y-axis the y- axis doesnt change but the x-axis changed.
Hope this help!

7 0

2 years ago

Read 2 more answers