If 2(x + 3) = x + 10, what is the value of x? O A 4 OB 14 O C. 5 O D.7 HURRY PLEASE

Mathematics

1 answer:

algol [13]3 years ago

4 0

Answer:

x=4

Step-by-step explanation:

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THERE ARE 3 BAGS OF APPLES WEIGHING A TOTAL OF 22 1/2 POUNDS. TWO OF THE BAGS WEIGH 6 3/8 POUNDS AND 3 1/4 POUNDS. HOW MUCH DOES

Mkey [24]

I wasn't going to click on this one, but the all-caps enthralled me and hypnotized me.

first off, let's change all the mixed fractions to "improper" and proceed from there, keep in mind that if we subtract the two bags' weight from the total, what's leftover is the 3rd bag's weight.

$\bf \stackrel{mixed}{22\frac{1}{2}}\implies \cfrac{22\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{45}{2}}
\\\\\\
\stackrel{mixed}{6\frac{3}{8}}\implies \cfrac{6\cdot 8+3}{8}\implies \stackrel{improper}{\cfrac{51}{8}}
\\\\\\
\stackrel{mixed}{3\frac{1}{4}}\implies \cfrac{3\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{13}{4}}\\\\
-------------------------------\\\\$

$\bf \stackrel{\textit{sum of the two bags}}{\cfrac{51}{8}+\cfrac{13}{4}}\impliedby \textit{our LCD is 8}\implies \cfrac{(1)51+(2)13}{8}
\\\\\\
\cfrac{51+26}{8}\implies \cfrac{77}{8}\\\\
-------------------------------\\\\
\stackrel{total}{\cfrac{45}{2}}~-~\stackrel{two~bags}{\cfrac{77}{8}}\impliedby \textit{our LCD is again 8}\implies \cfrac{(4)45-(1)77}{8}
\\\\\\
\cfrac{180~~-~~77}{8}\implies \cfrac{103}{8}\implies \stackrel{third~bag}{12\frac{7}{8}}$

3 0

4 years ago

Dylan owes half as much money as he used to owe. If he used to owe $28, which of the following expressions would reflect how muc

IrinaVladis [17]

Answer:

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Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Multiply the sum of 7/8 and 3/4 by the differences of 5/36 from 11/13

Anni [7]

Answer:

$=-\frac{331}{288}=-1.14930\dots$

Step-by-step explanation:

$(\frac{7}{8} + \frac{3}{4} )\times (\frac{5}{36} - \frac{11}{13} )\\\\\mathrm{Join}\:\frac{7}{8}+\frac{3}{4}:\quad \frac{13}{8}\\\\=\frac{13}{8}\left(\frac{5}{36}-\frac{11}{13}\right)\\\\\mathrm{Join}\:\frac{5}{36}-\frac{11}{13}:\quad -\frac{331}{468}\\\\=\frac{13}{8}\left(-\frac{331}{468}\right)\\\\=-\frac{13}{8}\times\frac{331}{468}\\\\\mathrm{Cross-cancel\:common\:factor:}\:13\\=\frac{1}{8}\times\frac{331}{36}\\\\=-\frac{1\cdot \:331}{8\times\:36}\\\\=-\frac{331}{288}$