Straight

Hello,

The base itself is ($3$) as an example and the exponent ($^3$) is the number that determines

how many times the base is multiplied e.g. like (${3}^{3}$), if you have understood it, feel free to tell me the answer.

Straight

Hello,

($5x-2y=16$) and ($15x-7y=51$),

($x=\frac{16}{5}+\frac{2}{5}y$) and ($15x-7y=51$),

($15(\frac{16}{5}+\frac{2}{5}y)-7y=51$),

($y=-3$),

$x=\frac{16}{5}+\frac{2}{5}*(-3)$,

($x = 2$),

$(x,y)=(2,-3)$,

let us make a rehearsal,

$5*2-2*(-3)=16$ and $15*2-7*(-3)=51$,

simplify now to:

$16=16$ and $51=51$,

thus results in a true statement, whereby it applies:

$(x,y)=(2,-3)$,

Straight

Hello,

of 3-digit numbers or 4-digit numbers or at all? :)
 

Straight

Hello,

it does because this post is only available for users,

not for guests, log-in if you wanna see the post.

Straight

I can't use emojis, but you still get a cookie! :)

In the video they used faculty $!$, it applies:

$0!=1$

Straight

No, try it yourself, what did you get?

Straight

Hello,

(${x}^{2}-4x+{y}^{2}-6y-36=25$),

($x^2-4x-y^2-6y=25+36$),

($x^2-4x+y^2-6y=61$),

($x^2-4x+?+y^2-6y=61+?$),

to solve the equation $x^2-4x+4=(x-2)^2$,

we are adding 4:

($x^2-4x+4+y^2-6y=61+4$),

factorise the expression using square of a binomial ($a^2-2ab+b^2=(a-b)^2$),

$(x-2)^2+y^2-6y=61+4$

$(x-2)^2+y^2-6y=65$,

$(x-2)^2+y^2-6y+?=65+?$,

to solve the equation $y^2-6y+9=(y-3)^2$,

we are adding 9:

$(x-2)^2+y^2-6y+9=65+9$,

factorise the expression using square of a binomial $a^2-2ab+b^2=(a-b)^2$,

$(x-2)^2+(y-3)^2=65+9$,

$(x-2)^2+(y-3)^2=74$,

The equation can be written in the form $(x-p)^2+(y-q)^2=r^2$

so that it corresponds to a circle with radius ($r=\sqrt{74}$) and centre ($2,3$).

Straight

Hello,

maybe you're not the same with him?

Or is this just a random number and would be a joke?

Straight

Hello,

the Guest that have answered the question is you

because he knows the solution that is not possible

to solve because there are no values for this question.

How does he know the answer and even be sure?

That's problably you and maybe the values are not visible to me,

which would be unlikely, or?

Straight

Hello melody,

couldn't B be 0 because 8*0 is 0 and the solution must end with 0?

Yes, I do know that the solution is B=5, but why not?

Straight