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2 years ago

9

What is the composition of transformations mapping ABC to A'B'C"? The first transformation is The second transformation is AABC

is AA'B"C".

1 answer:

spin [16.1K]2 years ago

3 0

Answer:

The first Transition is a reflection over the line M

The second transition is a 180 degree rotation.

Step-by-step explanation:

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Combine as indicated by the signs. 8-y/3y + y+2/9y - 2/6y

Taya2010 [7]

Answer:

26 + y

----------

9y

Step-by-step explanation:

Your using parentheses here would remove a great deal of ambiguity. Looking at your 8-y/3y + y+2/9y - 2/6y, I have interpreted it to mean:

(8-y)/3y + (y+2)/9y - (2/6)y. For example, without parentheses, your 8-y/3y might be interpreted differently, as 8 - y/(3y), or 8 - 1/3.

Looking at (8-y)/3y + (y+2)/9y - (2/6)y again, we see three different denominators: 3y, 9y and 6 y. The LCD here is 9y. Multiplying all three terms of (8-y)/3y + (y+2)/9y - (2/6)y by the LCD, we get:

3(8-y) + (y+2) + 3y. We must now divide this by the LCD:

3(8-y) + (y+2) + 3y

--------------------------

9y

Next we need to perform the indicated multiplication:

24 - 3y + y + 2 + 3y

----------------------------

9y

and then to combine like terms:

24 + 2 - 3y + y + 3y, 26 + y

---------------------------- or -----------

9y 9y

7 0

2 years ago

Read 2 more answers

How can I solve this problem (5^3)^-2? Can you help me

ICE Princess25 [194]

First multiply the power, 3 x -2 = -6
Thus, 5^-6
Make the power positive
1/5^6
1/15625

4 0

3 years ago

Do 6-12 please its due in 15 minutes

LUCKY_DIMON [66]

Answer:

-6

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using <u>integration by parts</u>:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x$

$\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:$

$\implies \displaystyle -2\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+\text{C}$

Divide both sides by 2:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}$

Rewrite in the same format as the given integral:

$\displaystyle \implies - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

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1 year ago

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valentinak56 [21]

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