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kirill [66]
3 years ago
11

Solve for x pls show work 3(1 + x) - 5 =3x - 2

Mathematics
2 answers:
erik [133]3 years ago
8 0

Answer:

x = 0

Step-by-step explanation:

3(1+x) - 5 = 3x - 2

3 + 3x - 5 = 3x - 2

3x - 3x = 5 - 2 - 3

0 = 0

GaryK [48]3 years ago
4 0
3(1+x) - 5 =3x-2
3+3x-5=3x-2
3x-2

Hope this helps!
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2.Ax + By + C = 0 or Ax + By = C.

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Given f(x) = -x - 3, find f(-6)
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Answer:

plug in -6

-(-6) - 3

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4 years ago
Solve for x and find the value of the unknown angle measures.
snow_tiger [21]

Answer:

x = 20

m = 92

n = 68

p = 92

Step-by-step explanation:

x = 20

2 + 3x = 62

-2           -2

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

   3x = 60

  -----   -----

    3       2

    x = 20

m = 92

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180 - 88 = 92

n = 68

The vertical angle is 68 degrees

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Vertical angle is m, which is 92.

Hope this helped.

5 0
3 years ago
How do you write 1.6 x 10^4 in standard form
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Answer:

16000

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4 years ago
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Determine the exact formula for the following discrete models:
marshall27 [118]

I'm partial to solving with generating functions. Let

T(x)=\displaystyle\sum_{n\ge0}t_nx^n

Multiply both sides of the recurrence by x^{n+2} and sum over all n\ge0.

\displaystyle\sum_{n\ge0}2t_{n+2}x^{n+2}=\sum_{n\ge0}3t_{n+1}x^{n+2}+\sum_{n\ge0}2t_nx^{n+2}

Shift the indices and factor out powers of x as needed so that each series starts at the same index and power of x.

\displaystyle2\sum_{n\ge2}2t_nx^n=3x\sum_{n\ge1}t_nx^n+2x^2\sum_{n\ge0}t_nx^n

Now we can write each series in terms of the generating function T(x). Pull out the first few terms so that each series starts at the same index n=0.

2(T(x)-t_0-t_1x)=3x(T(x)-t_0)+2x^2T(x)

Solve for T(x):

T(x)=\dfrac{2-3x}{2-3x-2x^2}=\dfrac{2-3x}{(2+x)(1-2x)}

Splitting into partial fractions gives

T(x)=\dfrac85\dfrac1{2+x}+\dfrac15\dfrac1{1-2x}

which we can write as geometric series,

T(x)=\displaystyle\frac8{10}\sum_{n\ge0}\left(-\frac x2\right)^n+\frac15\sum_{n\ge0}(2x)^n

T(x)=\displaystyle\sum_{n\ge0}\left(\frac45\left(-\frac12\right)^n+\frac{2^n}5\right)x^n

which tells us

\boxed{t_n=\dfrac45\left(-\dfrac12\right)^n+\dfrac{2^n}5}

# # #

Just to illustrate another method you could consider, you can write the second recurrence in matrix form as

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By substitution, you can show that

\begin{bmatrix}y_{n+2}\\y_{n+1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}^{n+1}\begin{bmatrix}y_1\\y_0\end{bmatrix}

or

\begin{bmatrix}y_n\\y_{n-1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}^{n-1}\begin{bmatrix}y_1\\y_0\end{bmatrix}

Then solving the recurrence is a matter of diagonalizing the coefficient matrix, raising to the power of n-1, then multiplying by the column vector containing the initial values. The solution itself would be the entry in the first row of the resulting matrix.

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