asinus

Alan, thank you for pointing out the conventional way.
I knew him, but I have learned with the help of some participants and radix, how to bring the graph in the Forum. I took the opportunity. It was fun! You, Alan, thank you again!

Greeting asinus :- )  !

Hello ballhopper!

{ y^2 + x^2 = 10

y+x = 4

$y^{2}+x^{2}=10$ 

$y+x=4$

$y=4-x$

$\left(4-x\right) ^{2}+x^{2}=10$

$16-8x+x²+x²=10$

$8-4x+x² =5$

$x²-4x+3=0$

$ax²+bx+c=0$

$a=1; \ b=-4; \ c=3$

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$x = {4 \pm \sqrt{16-12} \over 2}$ 

$x_{1} = {4 + \sqrt{16-12} \over 2}=3$

$x_{2} = {4 - \sqrt{16-12} \over 2}=1$

$solution\ set = \{3 ;1\}$

Greeting asinus :- )  !

Hello ballhopper!

Give the coordinates of the circles center and its radius

(x-7)^2 + (y+4)^2 = 16

$y= \sqrt{16- \left(x-7\right) ^{2}} -4$

coordinates of the circles:  7 / - 4

r = 4

Greeting asinus :- )  !

find x if h(x) =10

$x \in \Re \ \ \ L=\Re$ 

$x\ can \ take \ any \ real \ number \ value.$

Hello Konzetsu!

I mistakenly entered at the end of x instead of p.
Please indulgence! Sorry!

16 - 3p = 2/3p + 5

$16 - 3p = 2/3p + 5$

$11 = 3p + 2/3p$

$11 = \frac{9\ p^{2}+ 2 }{3p}$

$33p = 9p^{2}+ 2$

$9p^{2}- 33p+ 2= 0$

$a= 9; b= -33;c= 2$

$p = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$p = {33 \pm \sqrt{33^2-4*9*2} \over 2*9}$  

$p_{1}= \frac{33+ \sqrt{1017} }{18} = 3.605$

$p_{2}= \frac{33- \sqrt{1017} }{18} = 0.06164$

Greeting asinus :- )  !

Hello Konzetsu!

16 - 3p = 2/3p + 5

$16 - 3p = 2/3p + 5$

$11 = 3p + 2/3p$

$11 = \frac{9\ p^{2}+ 2 }{3p}$

$33p = 9p^{2}+ 2$

$9p^{2}- 33p+ 2= 0$

$a= 9; b= -33;c= 2$

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$x = {33 \pm \sqrt{33^2-4*9*2} \over 2*9}$  

$x_{1}= \frac{33+ \sqrt{1017} }{18} = 3.605$

$x_{2}= \frac{33- \sqrt{1017} }{18} = 0.06164$

Greeting asinus :- )  !

Guten Morgen lieber Gast!

Was soll überschlagen werden?

Bitte gib uns schnell einen Tip, worum es in deinem Test geht.

Dann kann dir sicher geholfen werden.

Gruß

asinus :- )

  !

-[6x-(4x+8)]=9+(6x+3)

$-[6x-(4x+8)]=9+(6x+3)$

$-[6x-4x-8]=9+6x+3$

$-6x+4x+8=9+6x+3$

$-8x=4$

$x=-\frac{1}{2}$

find x if h(x) =10

$x \in \Re$

find x if h(x) =10

$x \in \Re$