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Daniel [21]
3 years ago
13

Deangelo has $7 worth of dimes and quarters in a jar. He has 7 more quarters than dimes.

Mathematics
2 answers:
Aleksandr [31]3 years ago
4 0
Deangelo has 15 dimes and 22 quarters

Hope this helps
svlad2 [7]3 years ago
3 0
D=number of dimes
q=number of quarters

let's count everything in cents
dimes are worth 10 cents
quarters are worth 25 cents

10d+25q=700
divide both sides by 5
2d+5q=140


he has 7 more quarters than dimes
q=7+d
subsitute 7+d for q in other equation

2d+5q=140
2d+5(7+d)=140
2d+35+5d=140
7d+35=140
minus 35 both sides
7d=105
divide both sides by 7
d=15

sub back

q=7+d
q=7+15
q=22


22 quarters and 15 dimes
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\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill


\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=\sqrt{(-3-6)^2+(10+1)^2}\implies DN=\sqrt{202}


now that we know how long each one is, let's plug those in Heron's Area formula.


\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18

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