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Mathematics

1 answer:

AysviL [449]3 years ago

4 0

Answer:

Option (b) second one is the correct answer

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Multiply the polynomials (5x^2+4x-4)(4x^3-2x+6)<br> Explain your steps!!

Lapatulllka [165]

So distribute using distributive property
a(b+c)=ab+ac so

split it up
(5x^2+4x-4)(4x^3-2x+6)=(5x^2)(4x^3-2x+6)+(4x)(4x^3-2x+6)+(-4)(4x^3-2x+6)=[(5x^2)(4x^3)+(5x^2)(-2x)+(5x^2)(6)]+[(4x)(4x^3)+(4x)(-2x)+(4x)(6)]+[(-4)(4x^3)+(-4)(-2x)+(-4)(6)]=(20x^5)+(-10x^3)+(30x^2)+(16x^4)+(-8x^2)+(24x)+(-16x^3)+(8x)+(-24)
group like terms
[20x^5]+[16x^4]+[-10x^3-16x^3]+[30x^2-8x^2]+[24x+8x]+[-24]=20x^5+16x^4-26x^3+22x^2+32x-24

the asnwer is 20x^5+16x^4-26x^3+22x^2+32x-24

4 0

3 years ago

Help please! I will mark Brainiest!!

aleksklad [387]

Answer:

It is equal to each other

Step-by-step explanation:

$\sqrt{\frac{1}{x^2}} = \frac{1}{x}\\\sqrt[3]{\frac{1}{x^3}} = \frac{1}{x}$

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2 years ago

Rosa had $275. This table shows the items she purchased from the mall.

Serjik [45]

Upload a pic so we can help

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2 years ago

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Natalija [7]

Answer:

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Step-by-step explanation:

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3 years ago

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Keenan scored 80 points on an exam that had a mean score of 77 points and a standard deviation of 4.9 points. Rachel scored 78 p

Artemon [7]

Answer:

Keenan's z-score was of 0.61.

Rachel's z-score was of 0.81.

Step-by-step explanation:

Z-score:

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.

Keenan scored 80 points on an exam that had a mean score of 77 points and a standard deviation of 4.9 points.

This means that $X = 80, \mu = 77, \sigma = 4.9$