The quotient and remainder when the first polynomial is divided by the second are -4w^2 - 7w - 21 and -71 respectively
<h3>How to determine the quotient and remainder when the first polynomial is divided by the second?</h3>
The polynomials are given as:
-4w^3 + 5w^2 - 8, w - 3
Set the divisor to 0.
So, we have
w - 3 = 0
Add 3 to both sides
w = 3
Substitute w = 3 in -4w^3 + 5w^2 - 8 to determine the remainder
-4(3)^3 + 5(3)^2 - 8
Evaluate the expression
-71
This means that the remainder when -4w^3 + 5w^2 - 8 is divided by w - 3 is -71
The quotient (Q) is calculated as follows:
Q = [-4w^3 + 5w^2 - 8]/[w - 3]
The numerator can be expressed as follows:
Numerator = -4w^3 + 5w^2 - 8
Subtract the remainder.
So, we have:
Numerator = -4w^3 + 5w^2 - 8 + 71
This gives
Numerator = -4w^3 + 5w^2 + 61
So, the quotient becomes
Q = [-4w^3 + 5w^2 + 61]/[w - 3]
Expand
Q = [(w - 3)(-4w^2 - 7w - 21)]/[w - 3]
Evaluate
Q = -4w^2 - 7w - 21
Hence, the quotient and remainder when the first polynomial is divided by the second are -4w^2 - 7w - 21 and -71 respectively
Read more about polynomials at:
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Which data set has an outlier? 25, 36, 44, 51, 62, 77 3, 3, 3, 7, 9, 9, 10, 14 8, 17, 18, 20, 20, 21, 23, 26, 31, 39 63, 65, 66,
umka21 [38]
It's hard to tell where one set ends and the next starts. I think it's
A. 25, 36, 44, 51, 62, 77
B. 3, 3, 3, 7, 9, 9, 10, 14
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Let's go through them.
A. 25, 36, 44, 51, 62, 77
That looks OK, standard deviation around 20, mean around 50, points with 2 standard deviations of the mean.
B. 3, 3, 3, 7, 9, 9, 10, 14
Average around 7, sigma around 4, within 2 sigma, seems ok.
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
Average around 20, sigma around 8, that 39 is hanging out there past two sigma. Let's reserve judgement and compare to the next one.
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Average around 74, sigma 8, seems very tight.
I guess we conclude C has the outlier 39. That one doesn't seem like much of an outlier to me; I was looking for a lone point hanging out at five or six sigma.
well, this is just a matter of simple unit conversion, so let's recall that one revolution on a circle is just one-go-around, radians wise that'll be 2π, and we also know that 1 minute has 60 seconds, let's use those values for our product.
![\cfrac{300~~\begin{matrix} r \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{~~\begin{matrix} min \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{2\pi ~rad}{~~\begin{matrix} r \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{~~\begin{matrix} min \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{60secs}\implies \cfrac{(300)(2\pi )rad}{60secs}\implies 10\pi ~\frac{rad}{secs}\approx 31.42~\frac{rad}{secs}](https://tex.z-dn.net/?f=%5Ccfrac%7B300~~%5Cbegin%7Bmatrix%7D%20r%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7B~~%5Cbegin%7Bmatrix%7D%20min%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%5Ccdot%20%5Ccfrac%7B2%5Cpi%20~rad%7D%7B~~%5Cbegin%7Bmatrix%7D%20r%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%5Ccdot%20%5Ccfrac%7B~~%5Cbegin%7Bmatrix%7D%20min%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7B60secs%7D%5Cimplies%20%5Ccfrac%7B%28300%29%282%5Cpi%20%29rad%7D%7B60secs%7D%5Cimplies%2010%5Cpi%20~%5Cfrac%7Brad%7D%7Bsecs%7D%5Capprox%2031.42~%5Cfrac%7Brad%7D%7Bsecs%7D)
Answer:
Slope is 1.5 and y-intercept is 11
Step-by-step explanation:
It's in the equation
