A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
I will medal!
Fifteen percent of the pigment in paint color A is black. Sixty percent of the pigment in paint color B is black. An unknown amount of paint color B is mixed with 40 ml of paint color A, resulting in a paint that contains 25% black pigment.
Which equation can be used to solve for x, the total amount of paint in the mixture of the two colors?
0.15(40) + 0.6x = 0.25(40 + x)
0.15(40) + 0.6(x – 40) = 0.25(x)
0.15(40) + 0.6x = 0.25(40 – x)
0.15(40) + 0.6(x + 40) = 0.25(x)
anonymous
 one year ago
I will medal! Fifteen percent of the pigment in paint color A is black. Sixty percent of the pigment in paint color B is black. An unknown amount of paint color B is mixed with 40 ml of paint color A, resulting in a paint that contains 25% black pigment. Which equation can be used to solve for x, the total amount of paint in the mixture of the two colors? 0.15(40) + 0.6x = 0.25(40 + x) 0.15(40) + 0.6(x – 40) = 0.25(x) 0.15(40) + 0.6x = 0.25(40 – x) 0.15(40) + 0.6(x + 40) = 0.25(x)

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It should be A. :) The two terms on the left side represent the amount of black pigment with each paint color. On the right side it has the 25% in the new color combination of A (40ml) and the unknown amount (x) of B

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[0.15(40) + 0.6x = 0.25(40 + x)\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.