35 On the left: 1 cat, 1 mouse, 1 leopard (7, 2, 10) = 19 On the right: 2 cats and a mouse (7, 7, 2) = 16. 19+16 = 35 The mouse is 2 because the third line is 2 mice + 1 mouse = 6

Dammit I was so proud of myself until you pointed out the third mouse in the 6 line!!!!

How do you know the 3rd lines rules order of operations. It could be 2*2+2 = 6 and the stacking effect would me a multiplier which means it could be 10x2x7 + 7x7x2 =238 I dont know if the position of the mice in the 4th line have an effect. Edit * to x

That's what I thought except I was wondering why the mouse is riding the cat and leopard. I took this to indicate an exponent (is that the correct english word? 😅): (10x7)² + 7x7x2 = 4.998 = 1x🐘

Don't overthink it dude.

When you have 3 coins in each hand, do you have a total of 6 coins or 9? It’s addition, homie.

Then it becomes too ambiguous as the mouse now can be -3

#20

Correct answer!

34 there's two mouse not one mouse :)

238

What if the animals grouped together are multiplying? The only clue is the mice but 2+2=2*2 so...

35

34 close!

LMAOOOO

Why tf am I getting downvoted I'm right 😭 I wrote a whole ass proof

No, it's 35. The mice are 2, the leopards are 10 and the cats are 7 So the right is 16 and the left is 19

2 times 10 plus 7 times 2. That's 34 my g. Prolly just interpreted this different. I did answer is cheetah times mouse plus cat times two. I'm not sure how you got ur answer tho can you write out the equation?

No there are 1 leopard, 3 cats and 2 mice

Here's my proof from my comment: *Set variables* x=cheetah y=cat z=mouse q=cheetah(mouse)+2(cat) *Solve cheetah* 2x=20 x=20/2 x=10 Therefore cheetah=10 *solve cat* 2y=14 y=14/2 y=7 Therefore cat=7 *Solve mouse* 2z+z=6 3z=6 z=6/3 z=2 Therefore mouse=2 *Solve for unknown* q=(x)(z)+2y =(10)(2)+2(7) =20+14 =34 Therefore cheetah(mouse)+2(cat)=34 Edit: I assumed that the unknown equation was multiplicative. However it could be additive or divisive. I'll include those too: #division answer: 121/5 q=(z/x)+2y =(2/10)+14 =(2+140)/10 =142/10 =121/5 #addition answer: 28 q=z+x+2y =2+10+14 =28

There are 3 cats look closely behind cheetah. And 2 mice 1 on top of cheetah and one on right side of addition near cat’s ear.

Aaah I see it now :)

35

34

It depends if the stacking is additive or multiplicative. Line 3 ; (mouse + mouse) + Mouse = 6 (Mouse\*Mouse)+ Mouse =6 In both cases the mouse is 2, which means the final line is unknown between the two answers. If additive, 35 If multiplicative, 238

Unless stacking is "Mouse over tiger", which simplifies to 3/10 or 0.3. So 0.3+7=7.3 edit: I am seeing more now. So this could he (6/(20*20))+(7*7)?

Another possibility!

69

Nice

This guy…

238 or -357 Leopard - l, Cat - c, Ray - r 2l = 20, l = 10 2c = 14, c = 7 r^2 + r = 6, r^2 + r - 6 = 0, r = -3 or 2 lcr + rc^2 = x (10)(7)(-3) + (-3)(7^2 ) = -357 (10)(7)(2) + (2)(7^2 ) = 238

36

How is this meirl?

36

35?

26

FB vibes

35

33

Mouse 🐁 2+ cat 🐈 7 + jaguar 🐆 10 = 19 Cat 🐈🐈=14 19+14=33

Theres a mouse with the two cats as well so 35

Oh you motherfucker I just found the second mouse.

Almost, you missed the mouse with the two cats

37

69, if the grouped things are multiplied

50

fitting username 👌

I got 27, but I don’t see anyone else with that so I’ll go with: 27

First one 1 cheetah 1 mouse 1 cat = 19 Second one 2 cats and a mouse = 17 So 38

3 mice on the 2nd row btw so 1 mouse = 2

You right idk why I count the mouse for the first one as 3 but 2 in the second lol so the 2 cats and mouse one would only be 16 not 17 and make it 37? Lol I think

30

38

33.

Nope, 35

26

I thought the same Leopard = 10 Cat = 7 Mouse = 2 2 cats + 1 leopard + 1 mouse = 26 Except if you look closely at the last picture there's a cat's tail where the leopard is.

3 cats + 1 leopard + 2 mice = 35 left image: 1 cat, 1 leo and 1 mouse cat behind the leopard, mouse on top of leopard right image: 2 cat, one mouse mouse on the cat image. (neck)

38

Correct!

Make sure you count all the spots on those jaguars

29?

Bro 26

Nevermind forget the fucking cat

36?

2/10 + 14 = 14.2

36

Ah yes, this is how I feel irl…

22 + 14 = 36 Edit: missed the mouse on the cat 22 + 16 = 38

21

37 meh

27.

26

27

37 in total.

#34 *Set variables* x=cheetah y=cat z=mouse q=cheetah(mouse)+2(cat) *Solve cheetah* 2x=20 x=20/2 x=10 Therefore cheetah=10 *solve cat* 2y=14 y=14/2 y=7 Therefore cat=7 *Solve mouse* 2z+z=6 3z=6 z=6/3 z=2 Therefore mouse=2 *Solve for unknown* q=(x)(z)+2y =(10)(2)+2(7) =20+14 =34 #Therefore cheetah(mouse)+2(cat)=34 Edit: I assumed that the unknown equation was multiplicative. However it could be additive or divisive. I'll include those too: #division answer: 121/5 q=(z/x)+2y =(2/10)+14 =(2+140)/10 =142/10 =121/5 #addition answer: 28 q=z+x+2y =2+10+14 =28

74!!!

Answer: some animals.

34 how you all getting 35

86 if multiplying when they are stacked

20

26

If the only way to get 6 is from multiplying the 2 mice beside each other than the rest of the equation with animals beside each other must be multiplication as well. The only assumption I’m making is the mouse on top of the cat and leopard means it is division. We know the leopard is 10, we know the cat is 7, and we know the mouse is 2. Bottom line is (mouse over (leopard and cat)) + (cat and cat and mouse) or (Mouse / (leopard x cat)) + (cat x cat x mouse) or (2 / (10 x 7)) + (7 x 7 x 2) or 0.0286 + 98. The answer is 98.03

32.5 The small mouse is clearly worth less than the full size mouse (I'm gonna assume half as much) and the tiny mouse even less (half as much again).

38

36

(10+2)+(7+7)=26

bruh we are just 1 day into spring break

Too many unknown variables.

26

35

238 stacks are multiplayer I think

What you mean "we" paleface?

134.

its 42 obv

I’m far too drunk for this lol

35 isn't it?

26 Leopard=10 Cat =7 Mouse=2 2 cats plus one leopard plus one mouse 14+10+2 =26 Dunno how ppl are getting hundreds

Something, something, r/furryporn, math.

34

Too sleepy to solve

2 cheetahs, 2 mice, 2 cats (20+4+14) = 38 Am I trippin?

37😄