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3 years ago

Lcm of (x-2)(x+3) and 10(x+3)^2

1 answer:

8 0

Least common multiple: factor them, then see what they have in common and what is leftover and multiply those expressions:

(x - 2)(x + 3) 10(x + 3)(x + 3)
Common: (x + 3)
Leftover: (x - 2), (10), (x + 3)
Common · Leftover is: (x + 3) · (x - 2) · (10) · (x + 3) = 10(x - 2)(x + 3)²

Answer: LCM is 10(x - 2)(x + 3)²

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First I'll simplify the first one by distributing
11m + 10p
And the second
10p + 11m
Yep they're equal

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4 years ago

Find x. <br> I need help please

Misha Larkins [42]

Answer:

$x = 4 \sqrt{3}$

Step-by-step explanation:

By geometric mean theorem and Pythagoras theorem:

${x}^{2} = { (\sqrt{8 \times 4} )}^{2} + {4}^{2} \\ \\ {x}^{2} = { (\sqrt{32} )}^{2} + {4}^{2} \\ \\ {x}^{2} = 32 + 16 \\ \\ {x}^{2} = 48 \\ \\ x = \sqrt{48} \\ \\ x = 4 \sqrt{3}$

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3 years ago

The inverse variation equation shows the relationship between wavelength in meters, x, and frequency, y.

Dmitrij [34]

The relationship between wavelength and frequency is expressed by the equation λν=c where λ is the wavelength, v is the frequency and c is the speed of light which has a value of 3x10^8 m/s.

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Hope this answers the question.

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4 years ago

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I arrive at a bus stop at a time that is normally distributed with mean 08:00 and SD 2 minutes. My bus arrives at the stop at an

Nimfa-mama [501]

Answer:

0.0485 = 4.85% probability that you miss the bus.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When two normal distributions are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In this question:

We have to find the distribution for the difference in times between when you arrive and when the bus arrives.

You arrive at 8, so we consider the mean 0. The bus arrives at 8:05, 5 minutes later, so we consider mean 5. This means that the mean is:

$\mu = 0 - 5 = -5$

The standard deviation of your arrival time is of 2 minutes, while for the bus it is 3. So

$\sigma = \sqrt{2^2 + 3^2} = \sqrt{13}$

The bus remains at the stop for 1 minute and then leaves. What is the chance that I miss the bus?

You will miss the bus if the difference is larger than 1. So this probability is 1 subtracted by the pvalue of Z when X = 1.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1 - (-5)}{\sqrt{13}}$

$Z = \frac{6}{\sqrt{13}}$

$Z = 1.66$

$Z = 1.66$ has a pvalue of 0.9515

1 - 0.9515 = 0.0485

0.0485 = 4.85% probability that you miss the bus.

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3 years ago