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
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}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B1%7D%7Bx%5E2%7D%7D%20%3D%20%5Cfrac%7B1%7D%7Bx%7D%5C%5C%5Csqrt%5B3%5D%7B%5Cfrac%7B1%7D%7Bx%5E3%7D%7D%20%3D%20%5Cfrac%7B1%7D%7Bx%7D)
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Answer:
I dont speak that
Step-by-step explanation:
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
and standard deviation
, the z-score of a measure X is given by:

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 
So



Keenan's z-score was of 0.61.
Rachel scored 78 points on an exam that had a mean score of 75 points and a standard deviation of 3.7 points.
This means that
. So



Rachel's z-score was of 0.81.