Ask your own question, for FREE!
Mathematics 23 Online
OpenStudy (anonymous):

Medal and fan. help asap

OpenStudy (anonymous):

Jeremiah is making a shake by mixing two different protein powders, measured in ounces. The strawberry-flavored powder has 4 grams of protein per ounce. The banana-flavored powder has 3 grams of protein per ounce. He wants the drink to have a total of 6 ounces of powder and contain 22 grams of protein. Create a system of linear equations to represent the situation that will determine the exact number of ounces needed for each power.

OpenStudy (anonymous):

@jigglypuff314

jigglypuff314 (jigglypuff314):

per ounce is like per 1 ounce so as for the ounces you would get x + y = 6 then grams plus grams = grams so 4x + 3y = 22 so those would be your two equations for your system of equations :)

OpenStudy (anonymous):

and then it ask me Based on the system you wrote, which algebraic method of solving systems of equations will you choose to solve the system? Explain your choice.

jigglypuff314 (jigglypuff314):

mmm since you could easily isolate y fron x + y = 6 I would personally use substitution :)

OpenStudy (anonymous):

and one more question i already got #1 which is a.

OpenStudy (anonymous):

Several systems of equations are given below. System 1 y = 6x – 1.5 y = –6x + 1.5 System 2 x + 3y = –6 2x + 6y = 3 System 3 2x –y = 5 6x – 3y = 15 Which system of equations is consistent-independent? How many solutions will the system of equations have? Expain your answers. Which system of equations is consistent-dependent? How many solutions will the system of equations have? Expain your answers. Which system of equations is inconsistent-independent? How many solutions will the system of equations have? Expain your answers.

jigglypuff314 (jigglypuff314):

erm, I never learned the consistent/inconsistent dependent stuff .-. sorry :/

OpenStudy (anonymous):

thanks

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!