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Mademuasel [1]
4 years ago
9

Can someone help me please?? in Algblra2 - Variations, Progression, and Theorems

Mathematics
1 answer:
VashaNatasha [74]4 years ago
5 0
If (y-1) is a factor of f(y), f(y)=0 when y=1.  So if you find that f(1)=0, then (y-1) is a factor of f(y).

f(y)=y^3-9y^2+10y+5

f(1)=1-9+10+5=7

Since f(1)=7, (y-1) is not a factor.
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Find the area of the region defined by the region defined by the inequality 2|x| + 3|y-1| ≤ 6
EleoNora [17]

If x and y-1 have the same sign, then either

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or

x

If x and y-1 have opposite sign, then

x>0,y

or

x1 \implies 2|x| + 3|y-1| = -2x + 3(y-1) = 6 \implies 2x-3y = -9

This is to say that the region has boundaries given by these two sets of parallel lines, so we can equivalently describe the region with the set

R = \left\{(x,y) \mid -3\le2x+3y\le9 \text{ and } -9\le2x-3y\le3\right\}

The area of R is given by the double integral

\displaystyle \iint_R dx\,dy

To compute the area, change the variables to

\begin{cases}u = 2x + 3y \\ v = 2x - 3y\end{cases} \implies \begin{cases}x = \frac14(u+v) \\ y = \frac16(u-v)\end{cases}

The Jacobian for this transformation is

J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix}1/4 & 1/4 \\ 1/6 & -1/6\end{bmatrix}

with determinant \det(J) = -\frac1{12}. Then the integral transforms to

\displaystyle \iint_R dx\,dy = \iint_R |J| \, du \, dv = \frac1{12} \int_{-3}^9 \int_{-9}^3 dv\, du

which is 1/12 the area of a square with side length 12. Hence the integral evaluates to

\displaystyle \iint_R dx\,dy = \frac1{12}\times12^2 = \boxed{12}.

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