Craft project

1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a

katovenus [111]

1. Let a and b be coefficients such that

$\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}$

Combining the fractions on the right gives

$\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}$

$\implies 1 = (2a+b)x + 3a$

$\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23$

so that

$\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}$

2. a. The given ODE is separable as

$x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}$

Using the result of part (1), integrating both sides gives

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C$

Given that y = 1 when x = 1, we find

$\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)$

so the particular solution to the ODE is

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)$

We can solve this explicitly for y :

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)$

$\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|$

$\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|$

$\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}$

2. b. When x = 9, we get

$y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}$

8 0

2 years ago

Provide an Example of how to draw transformed figures that are translated, reflected, and rotated.

Paladinen [302]

A coordinate grid is very handy when it comes to drawing geometric shapes such as triangles. Let's create an example triangle ABC with the locations

A = (2,3)

B = (9,5)

C = (4,-10)

Plot those points and connect the dots. That forms triangle ABC. We can translate triangle ABC to any other position we want. Let's say we want to shift it 2 units to the left. That means we subtract 2 from each x coordinate while keeping the y coordinates the same. Therefore

A' = (0, 3)

B' = (7, 5)

C' = (2,-10)

Plot triangle A'B'C' and you should see that this is a shifted copy of triangle ABC.

The rotation rules are a bit more complicated, and it depends where you place the center of rotation; however, it is possible to use coordinate math like done above.

Luckily the reflection rules over the x or y axis are fairly simple. If we reflect over the x axis, then we flip the sign of the y coordinate. Or if we wanted to reflect over the y axis, we flip the sign of the x coordinate.

Example: A' = (0,3) reflects over the x axis to get A'' = (0, -3)

7 0

3 years ago

Please help. thank so much !!

DanielleElmas [232]

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2

Put it like that and there’s your constant

4 0

3 years ago

If 5y + 3 = 2y + 9 then find 2y=

Serhud [2]

If you were to solve for y, it would be 2. Therefore 2y would be 4.

5 0