Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 187595 by Ari last updated on 19/Feb/23

Commented by Ari last updated on 19/Feb/23

Who can help me with problem please!I can not create a function in this situation

Answered by mr W last updated on 19/Feb/23

x=number of tables of type 1  y=number of tables of type 2  c=number of customers  a)  c=4x+6y  this is a function between x, y and c.  b)  c=4x+6y=50  ⇒2x+3y=25  ⇒2x=25−3y  we see y must be odd such that 25−3y  is even.  0≤25−3y   ⇒y≤8 ⇒y=1,3,5,7.  that means there are 4 possibilities:  1 table of type 2 and 11 tables of type 1,  3 tables of type 2 and 8 tables of type 1,  5 tables of type 2 and 5 tables of type 1,  7 tables of type 2 and 2 tables of type 1.

x=numberoftablesoftype1y=numberoftablesoftype2c=numberofcustomersa)c=4x+6ythisisafunctionbetweenx,yandc.b)c=4x+6y=502x+3y=252x=253yweseeymustbeoddsuchthat253yiseven.0253yy8y=1,3,5,7.thatmeansthereare4possibilities:1tableoftype2and11tablesoftype1,3tablesoftype2and8tablesoftype1,5tablesoftype2and5tablesoftype1,7tablesoftype2and2tablesoftype1.

Commented by Ari last updated on 19/Feb/23

Thank you very much sir!

Terms of Service

Privacy Policy

Contact: info@tinkutara.com