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3 years ago

If f (x)=|3x-2|. Then what is f(5)

Mathematics

1 answer:

vichka [17]3 years ago

6 0

Answer:

Step-by-step explanation:

|3(5)-2|

|15-2|

|13|

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What is the solution to the equation?

AlekseyPX

Answer:

C) -3

Step-by-step explanation:

$-\frac{5x}{3}$ + 8 = 13

$-\frac{5x}{3}$ = 5

-5x = 15

x = -3

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3 years ago

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4x + 3y = 8<br>2x - y=9

Alex17521 [72]

Step-by-step explanation:

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Can someone help me with this problem?

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5/7, I don't know what that is as a decimal, but I hope it helps

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4 years ago

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What are the lengths of the legs of a right triangle in which one acute angle measures 19° and the hypotenuse is 15 units long?

Deffense [45]

So you have a right angle triangle. You need to use trig for this, as you have two values you can figure out a third.

You know that you have

90 deg. angle
19 deg. angle
Therefor you have 71 deg. angle

sin19 = x/15
15sin19 = x
x = 2.2

Now do Pythagoras c^2-b^2=a^2

a = 14.8

So the sides are 2.2 and 14.8 units long. (If these numbers are wrong tell me and I'll edit answer)

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3 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,

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

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

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

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

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3 years ago