All categories

4 years ago

Which of the following circles have their centers in the second quadrant? Check all that apply.

Mathematics

1 answer:

steposvetlana [31]4 years ago

3 0

B. (x + 1)2 + (y - 7)2 = 16
D. (x + 2)2 + (y - 5)2 = 9

You might be interested in

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

Find an equation of the tangent line to y = 2 sin x at x = _/6

lys-0071 [83]

Y o = 2 sin ( π/6 ) = 2 * 1/2 = 1
x o = π / 6
y ` = 2 cos x = 2 cos (π/6) = 2 * √3 / 2 = √ 3;
m = √3
An equation of the tangent line is:
y - y o = m ( x - x o )
Answer:
y - 1 = √3 ( x - π/6 )

5 0

4 years ago

1. Jovi and Amillia are playing a game using the spinner below. Jovi will win the game on his next spin if the arrow lands on a

Lubov Fominskaja [6]

For question number one, Jovi WILL most likely win.

Hope this helps.

8 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! What is (f−g)(x)?

Korolek [52]

Answer: 3x⁵ - 2x⁴ - x² + x - 21

Step-by-step explanation:

f(x) = 3x⁵ + 6x² - 5

- g(x) = -2x⁴ - 7x² + x - 16

(f - g)(x) = 3x⁵ - 2x⁴ - x² + x - 21

3 0

3 years ago

A $300 phone is on sale for 20% and you have a coupon for an extra 10% off. How much will you pay for the phone? (no tax) ?

kkurt [141]

Answer:

You will pay $210 for the phone

Step-by-step explanation:

20%+10%= 30%

100%-30%= 70%

70% of $300 is $210

5 0

3 years ago