why isn't there the same outcome to both calculuses? am

>both calculuses
OP this is simple multiplication of fractions, not calculus. Stick to brainlet activities.

>>9116351
>retarded

2 = 8/4
3 = 9/3

>>9116351
The second line those fractions are not equal to original numbers. you have to multiple the whole number by the denominator and add it to the numerator.

>>9116351
You are proof for humanity being doomed.
This level of stupidity is mindblowing for various reasons.

>>9116409
i'm 7
chill

3*2 = 6
So how can multiplying higher values equal less?
Just think rationally
Not like brainwashed dumbass

>>9116415
>think [math]\mathbb{Q}[/math]-ly
kek'd

>>9116413
Underageb&

Why should there be?
The bottom one isn't valid math.

>>9116351
9/4 * 11/3 = 8.25 is correct
It's a mystery how you got to it tho

The top line assumes [math]2\frac{1}{4}\cdot3\frac{2}{3}=(2+\frac{1}{4})\cdot(3+\frac{2}{3})[/math]

The bottom line assumes [math]2\frac{1}{4}\cdot3\frac{2}{3}=2\cdot\frac{1}{4}\cdot3\cdot\frac{2}{3}[/math]

The top line is the obviously correct interpretation.

>>9116413
[math]
2\frac{1}{4}\cdot 3\frac{2}{3} = \left (2+ \frac{1}{4} \right ) \left (3+\frac{2}{3} \right )
= \left (\frac{8}{4}+ \frac{1}{4} \right ) \left (\frac{9}{3}+\frac{2}{3} \right )
=\frac{9}{4}\cdot \frac{11}{3} = \frac{99}{12} = \frac{33\cdot 3}{4\cdot 3}=\frac{33}{4}=\frac{32}{4}+\frac{1}{4}=8+\frac{1}{4}=8\frac{1}{4}
[/math]

>>9116351
>calculuses
*tips fedora*

>>9116858
Calcula

>>9116351
In what planet 33/4 = 8/4, i definitely want to go there

>>9116365
Calculus can mean "calculation." See "impact calculus"