Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

If a bag of coins weighing 225 grams is filled with \(p\) pennies, \(n\) nickels, and \(d\) dimes, which of the following expresses \(d\) in terms of \(n\) and \(p\) ? A) \(100-\frac{10}{9}(p+2 n)\) B) \(100+\frac{10}{9}(p+2 n)\) C) \(100-\frac{10}{9}(p-2 n)\) D) \(100+\frac{10}{9}(p-2 n)\)

Short Answer

Expert verified
The short answer based on the given step-by-step solution is: \[d = 100-\frac{10}{9}(p+2n) \longrightarrow \boxed{\text{A}}\]

Step by step solution

01

Remove the Coin Units #

Since we want to obtain \(d\) in terms of \(n\) and \(p\), we should rewrite the equation in terms of these variables. To make this easier, we will eliminate the units from our equation by multiplying each term by the conversion factor to represent the number of each coin: \[225 (4.4) = 2.5(4.4)p + 5(4.4)n + 2.268(4.4)d \] Simplifying: \[990 = 11p + 22n + 10d \]
02

Isolate 'd' Variable #

Now, we will isolate \(d\) by putting all other terms on the right side of the equation: \[10d = 990 - 11p - 22n\] And divide each side by 10: \[d = 99 - \frac{11}{10}p - \frac{22}{10}n\]
03

Simplify d in Terms of n and p #

Finally, we can use common denominators and simplify the expression further: \[d = 99 - \frac{11}{10}p - \frac{11}{5}n\] \[d = 99 - \frac{11}{10}(p + 2n)\] Comparing this expression to the given alternatives, we find that the correct answer is: \[d = 100-\frac{10}{9}(p+2n) \longrightarrow \boxed{\text{A}}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free