All categories

3 years ago

What is the equation of a circle with the center (2,3) and radius 3

Mathematics

1 answer:

ruslelena [56]3 years ago

4 0

Answer:

(x - 2)² + (y - 3)² = 9

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

here (h, k) = (2, 3) and r = 3, hence

(x - 2)² + (y - 3)² = 9 ← equation of circle

You might be interested in

Derive Euler's formula e^(ix) - cos(x) + i*sin(x) by using taylor series expansion

gavmur [86]

Answer:

the answer is probably 5

5 0

2 years ago

Two friends start a business. Together they spent 800 dollars for their initial inventory. They will sell all of their initial i

Paul [167]

Their selling price was 2.45 times the price they bought it. If they split the profit evenly and there was two of them, multiply their profit by two and divide that number by 800 to find how much more they sold it for, so they sold it for 245% more than they bought it

5 0

3 years ago

The average age of the six people on a winter ski trip was twenty-six. An eleven year-old had to leave the trip early to attend

guapka [62]

It’s will be 25 or 36 ?

3 0

2 years ago

Three hamburgers and two hotdogs cost $13. Two hamburgers and four hotdogs cost $14 what is the cost of one hamburger? What is t

Harrizon [31]

Step-by-step explanation:

Let's take cost of 1 hamburger as h and the cost of 1 hot dog as d.

Therefore, we can form the equations:

3h + 2d = 13 and

2h + 4d = 14

2(h +2d) = 14

h + 2d = 7

h = 7 - 2d

By substituting this to the first equation, we get:

3(7 - 2d) + 2d = 13

21 - 4d = 13

4d = 8

d = $2

h = 7 - 2(2)

h = $3

Hence <u>1 hot dog costs $2</u> and <u>1 hamburger costs $3</u>

5 0

3 years ago

Use Cramer Rule to solve the following system: 8x−5y=70 and 9x+7y=3

nlexa [21]

Answer:

$(x,y) = (5,-6)$

Step-by-step explanation:

$\underline{\textbf{Determinant of a matrix.}}\\\\\text{For a}~ 2 \times 2 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2\\b_1&b_2 \end{vmatrix} = a_1b_2 - a_2b_1\\\\\\\text{For a}~ 3 \times 3 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2&b_3\\c_2&c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1&b_3\\c_1&c_3 \end{vmatrix}+ a_3 \begin{vmatrix} b_1&b_2\\c_1&c_2 \end{vmatrix}\\\\\\$

$~~~~~~~~~~~~~~~~~~=a_1(b_2c_3-b_3c_2) -a_2(b_1c_3-b_3c_1) +a_3(b_1c_2-b_2c_1)$

$\underline{\textbf{Cramer's Rule to solve a system of two equations.}}\\\\\text{Consider the system of two equations:}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_1x + b_1 y= c_1\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_2x +b_2 y = c_2\\\\\text{Here,}\\\\x = \dfrac{D_x}{D}= \dfrac{\begin{vmatrix} c_1&b_1\\c_2&b_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\\\ y= \dfrac{D_y}{D}= \dfrac{\begin{vmatrix} a_1&c_1\\a_2&c_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\$

$\underline{\textbf{Solution:}}\\\\~~~~~~~~~~~~~~~~~~~~~~~8x-5y = 70~~~~~~...(i)\\\\~~~~~~~~~~~~~~~~~~~~~~~9x +7y = 3~~~~~~~...(ii)\\\\\text{Applying Cramer's rule:}\\\\x = \dfrac{D_x}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 70& -5 \\3&7 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{70(7) -(-5)(3)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{490+15}{56+45}\\\\\\~~=\dfrac{505}{101}\\\\\\~~=5$

$y = \dfrac{D_y}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 8& 70 \\9&3 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{(8)(3) -(70)(9)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{24-630}{56+45}\\\\\\~~=-\dfrac{606}{101}\\\\\\~~=-6$

$\textbf{Hence, the solution to the system of equation is}~ (x,y) = (5,-6)$

7 0

1 year ago