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Fudgin [204]
3 years ago
9

I giveeeeee brainlilst

Mathematics
1 answer:
frosja888 [35]3 years ago
6 0

Answer:

C' (-9 , 3)  D' (6 , 3)  E' (-9 , 6)

Step-by-step explanation:

Dilation Factor: 3           Origin: (0,0)       (x,y) ---> (3x,3y)

C' (-9 , 3)

D' (6 , 3)

E' (-9 , 6)

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 A horizontal translation and a 180 rotation.
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Write the equation that passes through the following points:<br><br> (5,0) and (−10,−5)
Vikentia [17]

Answer:

The answer is:

y =  \frac{1}{3} x -  \frac{5}{3}

Step-by-step explanation:

❃Incase you forgot what the linear equation formula is ☟

y = mx + b

❃Incase you also forgot, The is what the slope formula is ☟

m = \frac{y_2 - y_1 }{x_2 - x_1}

➊ First: We are going to be solving for the slope.

m = \frac{  - 5 - 0}{ - 10 - 5} = \frac{ - 5}{  - 15} =  \frac{1}{3}

➋Second: We find the y-intercept.

y = mx + b \\ 0 =  \frac{1}{3} (5) + b \\ 0 =  \frac{5}{3}  + b \\ \frac{  -   \frac{5}{3}  =  -  \frac{5}{3 }   \:  \:  \:  \:  \:  \:  \: \: \: \: \: \: }{  -  \frac{5}{3}  = b}

➌Third: Plug in.

y =  \frac{1}{3} x -  \frac{5}{3}

4 0
3 years ago
Enter the missing values in the area model to find –3(5.5n - 1) 5.5n -1 -3 According to the model above, -3(5.5n - 1) ​
Elena-2011 [213]

Answer:

“Information science is the science and practice dealing with the effective collection, storage, retrieval, and use of information. It is concerned with recordable information and knowledge, and the technologies and related services that facilitate their management and use.

Step-by-step explanation:

3 0
2 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bp%7D%7B5%7D" id="TexFormula1" title="\frac{p}{5}" alt="\frac{p}{5}" align="absmiddle
tatyana61 [14]

Answer:

p ≥ -10

Step-by-step explanation:

first, subtract one from each side to get:

p/5 ≥ -2

lastly, multiply each side by 5

p ≥ -10

6 0
2 years ago
Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n
aliya0001 [1]

Answer:

f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...

We first compute the n-th derivative of f(x)=\ln(1+2x), note that

f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\

Now, if we compute the n-th derivative at 0 we get

f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!

and so the Maclaurin series for f(x)=ln(1+2x) is given by

f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2  \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}

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