/sci/ - Science & Math

[Anonymous No.9404352]

Solve for x, y and z.

[Anonymous No.9404358]

>>9404352
all 5

[Anonymous No.9404393]

>>9404352
they're all 5. you can't have negative length, brainlets.

[Anonymous No.9404407]

>>9404393
VECTORS

[Anonymous No.9404421]

>>9404358
>>9404393
>Thinking the problem is asking for the length of the hypotenuse.
/sci/ is full of brainlets. There's an obvious way to find out X, Y, and Z given the other values (let's denote vertical as A and horizontal as B).
Then we have [math]-\frac{1}{12}(3A+4B)\cdot A+\frac{A}{B}-\frac{B}{2}[/math]
This is a simple relationship that I could expect even my dumbest student to grasp. How it can even escape you brainlets will keep me up at night.

[Anonymous No.9404425]

>>9404352

All I see anymore are four jews who are clasping their hands in gleeful greed, antcipating possibly either the fruition of a lengthy plan which they have laid, or else possibly a simpler pleasure at the yield of some profit.

[Anonymous No.9404426]

>>9404393
>You can't have negative length
Depends on the situation. In this case, I agree with you

[Anonymous No.9404430]

>>9404407
Vector triangle are still calculated by norms, which are non-negative. Brainlet

[Anonymous No.9404469]

>>9404421
>5 == -7.5
nice

[Anonymous No.9404500]

>>9404352
They're all 5, bring on the next one. In this case, you can't have negative length ye brainlets.

[Anonymous No.9404518]

By definitions, distances in metric spaces are positive.

[Anonymous No.9404524]

>>9404352
-3 squared is the same as 3 squared.

[Anonymous No.9404525]

>>9404352
>>9404358
They're all -5 actually.

[Anonymous No.9404527]

>>9404469
Every day, the beauty of math finds another way to steal my breath.

[Anonymous No.9404532]

x = y = +/- 5i
z = -5

[Anonymous No.9404709]

[Anonymous No.9404755]

>>9404709
what does this mean

[Anonymous No.9404802]

>>9404755
i^2 +1^2 = 0^2 brainlet, prove me wrong