The answer would be 16 S'mores and the limiting reactant would be the grahams.
(This is assuming that S'mores would need 2 grahams, 1 marshmallow and 1 chocolate piece.)
Limiting reactant would be the reactant that runs out first.
Let's take your problem into account and see what we have:
48 marshallows
32 grahams (2 x 16 per pack)
45 chocolate pieces (5 x 15 pieces per bar)
Since need 2 of the grahams per S'more then the maximum yield of the grahams is 16 S'mores.
The maximum yield of marshmallows is 48.
The maximum yield of chocolate is 45.
Since you cannot make S'mores without the grahams, then you can only make 16 S'mores before the grahams run out.
Its C! Lets say x=30
|30-323|=50
This could technically be true because it would show if the temp is within range
Some of the possible points using this equation are...
(2,14)
(1,10)
(0,6)
(-1,2)
(-2,-2)
Answer:


Step-by-step explanation:
El cos(α) se define como el cociente entre el cateto adyacente y la hipotenusa.
El valor del cateto adyacente en nuestgro caso es CA = 3.
La hipotenusa se calcual de la siguiente manera:

Por lo tanto, el cos(α) sera:

El cosec(α)=h/CO.
El cateto opuesto CO = 4 y la hipotenusa h = 5
Por lo tanto, el cosec(α) sera:

Espero te haya sido de ayuda!
<span>F(x) = (1/4)^x
</span><span>F(3) = (1/4)^3 = 1/64
</span><span>B. 1/64</span>