Answer:
6\left(\dfrac{2}{3}+4\dfrac{1}{2}m\right)=6296(
3
2
+4
2
1
m)=6296, left parenthesis, start fraction, 2, divided by, 3, end fraction, plus, 4, start fraction, 1, divided by, 2, end fraction, m, right parenthesis, equals, 629 t
Step-by-step explanation:
i got this from khan
left parenthesis, start fraction, 2, divided by, 3, end fraction, plus, 4, start fraction, 1, divided by, 2, end fraction, m, right parenthesis, equals, 629 t
We have given that,
A display has 666 packs of marbles with a total mass of 629\,\text{g}629g629, start a text, g, end text. The packaging of each pack has a mass of \dfrac{2}{3}\,\text{g} 3 2 start fraction, 2, divided by, 3, end fraction, start a text, g, end text, and each marble has a mass of 4\dfrac{1}{2}\,\text{g}4 2 1 g4, start fraction, 1, divided by, 2, end fraction, create a text, g, end text.
Total mass means the mass of the shell, its service equipment, and structural equipment, and the heaviest load authorized to be transported.
We have to determine which equation can we use to determine mmm, the number of marbles per pack.
6\left(\dfrac{2}{3}+4\dfrac{1}{2}m\right)=6296(32+421m)=6296,
left parenthesis, start fraction, 2, divided by, 3, end fraction, plus, 4, start fraction, 1, divided by, 2, end fraction, m, right parenthesis, equals, 629 t.
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