$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{-3}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-3-1}{2-3}\implies \cfrac{-4}{-1}\implies 4$

8 0

2 years ago

System of conics help

ZanzabumX [31]

Answer:

Solutions are (6,3) and (-2,3).

Step-by-step explanation:

From the first equation given we can write:

$(x-2)^{2} = 16-(y-3)^{2}\\$

substituting for $(x-2)^{2}$ in the second equation given we get

$\frac{16-(y-3)^{2}}{16} + \frac{(y-3)^{2}}{64} = 1$

$\frac{-(y-3)^{2}}{16} + \frac{(y-3)^{2}}{64} =0$

∴ y=3

Putting y=3 in the first equation we get

$(x-2)^{2} = 16$

x-2 = ±4

Hence x=6 or x=-2.

7 0

3 years ago

Solve the system using multiplication for the linear combination method.

zysi [14]

(2,3) is your answer for the equation

5 0

3 years ago

Plz help as many times as i try i still don't get it

shepuryov [24]

I do know the bottom one is 2.8
Top is 1.6
I don’t know the middle one

3 0

2 years ago

Read 2 more answers