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Ksju [112]
3 years ago
14

together, tom and max have 72 football cards, tom has 2 more than 4 times as many cards as max has, how many football cards does

tom have?
Mathematics
1 answer:
andrew11 [14]3 years ago
7 0

Answer:

58

Step-by-step explanation:

Let t and m represent the numbers of cards had by Tom and Max, respectively. Then the problem statement tells us ...

  • t + m = 72
  • 4m +2 = t

We want to solve for t. If we multiply the first equation by 4 and subtract the second, we get ...

... 4(t +m) -(4m +2) = 4(72) -(t)

... 4t +4m -4m -2 = 288 -t . . . . . . . eliminate parentheses

... 5t = 290 . . . . . . . . . . . . . . . . . . . add t+2, collect terms

... t = 58 . . . . . . . . divide by the coefficient of t

_____

<em>Alternate Solution</em>

You can work this in your head if you realize that subtacting 2 from the total means that Max has 1/5 of the new amount. 1/5 of 70 is 14, so Tom has 72 - 14 = 58 cards.

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4 0
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Calculator
Sedaia [141]

Check the picture below.

so let's find the lengths of those two sides in red, since are the length and width of the rectangle.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{6})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[-3-(-6)]^2+[6-3]^2}\implies d=\sqrt{(-3+6)^2+(6-3)^2} \\\\\\ d=\sqrt{9+9}\implies \boxed{d=\sqrt{18}} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-1})~\hfill d=\sqrt{[-2-(-6)]^2+[-1-3]^2} \\\\\\ d=\sqrt{(-2+6)^2+(-1-3)^2}\implies d=\sqrt{16+16}\implies \boxed{d=\sqrt{32}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the rectangle}}{(\sqrt{18})(\sqrt{32})}\implies \sqrt{18\cdot 32}\implies \sqrt{576}\implies 24

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3 years ago
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