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4 years ago

Help plz...........................

Mathematics

2 answers:

stich3 [128]4 years ago

4 0

Answer:So its simple just forget about the letters add the numbers then place the letter and x

Step-by-step explanation: ( Always Parentheses first) Then i think you divide it by the outside number

lyudmila [28]4 years ago

4 0

Answer:

in order of the picture:

18w+8 -48x-24

16d+20 -48p-18

6v-16 25m-10

-24z+16 -9n+36

10d-2 -36k-36

-14j-10 -9b+6

24j-9 3v+5

-72v-16 18h-72

-20f+32 -7x+8

4p-6 45t+72

Step-by-step explanation:

Hope this helps.

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Natalie borrowed money from her parents to pay for a trip

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Answer: wow nice

Step-by-step explanation:

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3 years ago

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Naya [18.7K]

Answer:

y=x+5+x+5?

Step-by-step explanation:

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4 years ago

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Let an = –3an-1 + 10an-2 with initial conditions a1 = 29 and a2 = –47. a) Write the first 5 terms of the recurrence relation. b)

zlopas [31]

We can express the recurrence,

$\begin{cases}a_1=29\\a_2=-47\\a_n=-3a_{n-1}+10a_{n-2}7\text{for }n\ge3\end{cases}$

in matrix form as

$\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}$

By substitution,

$\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}\implies\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^2\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}$

and continuing in this way we would find that

$\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}\begin{bmatrix}a_2\\a_1\end{bmatrix}$

Diagonalizing the coefficient matrix gives us

$\begin{bmatrix}-3&10\\1&0\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}$

which makes taking the $(n-2)$ -th power trivial:

$\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}^{n-2}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}$

$\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}$

So we have

$\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}\begin{bmatrix}a_2\\a_1\end{bmatrix}$

and in particular,

$a_n=\dfrac{29\left(2(-5)^{n-1}+5\cdot2^{n-1}\right)-47\left(-(-5)^{n-1}+2^{n-1}\right)}7$

$a_n=\dfrac{105(-5)^{n-1}+98\cdot2^{n-1}}7$

$a_n=15(-5)^{n-1}+14\cdot2^{n-1}$

$\boxed{a_n=-3(-5)^n+7\cdot2^n}$

6 0

3 years ago

Point D' is the image of D(-2, 1) under a reflection across the x-axis.

makkiz [27]

Answer:

-2, -1

Step-by-step explanation:

the x same cause if you reflect across the x-axis so you move on the y-axis. and reflect means the same distance from the point you reflect (that it was 1 cause it was 1 point above the x-axis) just negative so -1 now. if the original point was -2,-1 so the answer was -2,1.

5 0

3 years ago

F(x) = 3 and g(x) = x – 2<br><br> 3 = x – 2<br><br> 5 = x

stiks02 [169]

Answer:

3=1×5-2

Step-by-step explanation:

you minus 5-2 is 3 and one threes three

4 0

2 years ago