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anygoal [31]
1 year ago
6

If the two lines below are perpendicular and the slope ofthe red line is -, what is the slope of the green line?51015-5-5-105

Mathematics
1 answer:
inysia [295]1 year ago
4 0

If two lines are perpendicular, then the product of their slopes is -1:

m_r\cdot m_g=-1...(1)

From the problem, we identify:

m_r=-\frac{4}{3}\text{  \lparen the slope of the red line\rparen}

Then, using the equation (1), we calculate the slope of the green line:

\begin{gathered} -\frac{4}{3}\cdot m_g=-1 \\  \\ \therefore m_g=\frac{3}{4} \end{gathered}

Answer: Option D

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<h3><u>(1 + 5y)(1 - 5y) is the fully factored form of this polynomial.</u></h3>

This polynomial can be factored using difference of squares.

This polynomial is in the form of a^2 - b^2 = (a + b)(a - b)

Because 1 * 1 = 1, we can use this formula to simplify this polynomial.

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How do you do this question?
Ksivusya [100]

Answer:

V = (About) 22.2, Graph = First graph/Graph in the attachment

Step-by-step explanation:

Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.

\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}

The plus 1 in '1 + 2/x' is shifting this graph up from where it is rotating, but the negative 1 is subtracting the area between the y-axis and the shaded region, so that when it's flipped around, it becomes a washer.

V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\

\int _1^3\left(1+\frac{2}{y}\right)^2dy=4\ln \left(3\right)+\frac{14}{3}, \int _1^31dy=2\\\\=> \pi \left(4\ln \left(3\right)+\frac{14}{3}-2\right)\\=> \pi \left(4\ln \left(3\right)+\frac{8}{3}\right)

Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.

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