The total mass of the marbles and packaging is equal to the total mass of the packs of marbles, so we can set up the equation:
\[(\text{mass of each marble})(\text{number of marbles per pack}) + (\text{mass of packaging}) = (\text{total mass of packs of marbles})\]
Substituting the given values, we get:
\[4\frac{1}{2}m + \frac{2}{3} = 629\]
So, the answer is $\boxed{\text{(A)} \: 6\cdot 4\frac{1}{2}m+\frac{2}{3}=629}$.
A display has
\[6\] packs of marbles with a total mass of
\[629\,\text{g}\]. The packaging of each pack has a mass of
\[\dfrac{2}{3}\,\text{g}\] and each marble has a mass of
\[4\dfrac{1}{2}\,\text{g}\].
Which equation can we use to determine
\[m\], the number of marbles per pack?
Choose 1 answer:
Choose 1 answer:
(Choice A)
\[6\cdot 4\dfrac{1}{2}m+\dfrac{2}{3}=629\]
A
\[6\cdot 4\dfrac{1}{2}m+\dfrac{2}{3}=629\]
(Choice B)
\[6m\left(4\dfrac{1}{2}+\dfrac{2}{3}\right)=629\]
B
\[6m\left(4\dfrac{1}{2}+\dfrac{2}{3}\right)=629\]
(Choice C)
\[6m+4\dfrac{1}{2}+\dfrac{2}{3}=629\]
C
\[6m+4\dfrac{1}{2}+\dfrac{2}{3}=629\]
(Choice D)
\[6\left(4\dfrac{1}{2}m+\dfrac{2}{3}\right)=629\]
D
\[6\left(4\dfrac{1}{2}m+\dfrac{2}{3}\right)=629\]
3 answers
so whats the answer a b c or d
The answer is $\boxed{\text{(A)} \: 6\cdot 4\frac{1}{2}m+\frac{2}{3}=629}$.
4 answers