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3 years ago

The measures of two supplementary angles are 12q - 9 and 8q + 14. Find the measures of the angles.

1 answer:

8 0

12q - 9 + 8q + 14 = 180
20q + 5 = 180
20q = 175
q = 8.75
angle 1 : 12(8.75) - 9 = 96
angle 2 : 8(8.75) + 14 = 84
Check: 84 + 96 = 180

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Ms. Lawrence had $550 in her bank account. Thr next day she withdrew $120 to buy groceries. She deposited $200 two days later. F

Feliz [49]

Answer:

$541

Step-by-step explanation:

1). $550 - $120 = $430 left in her bank account.

2). $430 + $200 = $630 in her bank account.

3). $630 - $89 = $541 left in her bank account.

(Withdraw means removing some of your money out of your bank account, and deposit means putting money back in your bank account)

So, therefore, after all of those transactions, Ms. Lawrence's bank account has $541.

Hope this helps! <3

6 0

4 years ago

Read 2 more answers

The lines represented by the equations 20y-24x=-20y and 5y-6x=-5 are

tankabanditka [31]

Answer:

The answer is perpendicular

Step-by-step explanation:

convert the two equations into slope-intercept form

y=7/5x - 1

y=6/5x -1

Then, put it in the graph and you'll get lines intersecting at the same point which is the y-intercept.

8 0

3 years ago

Read 2 more answers

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

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