Answer:
(10/3, 2/3)
Step-by-step explanation:
You may find it easier to write these two equations x+y=4 2x-y=6 in a column:
x+y=4
2x-y=6
Adding these together will eliminate y:
3x = 10. Then x = 10/3.
Substituting 10/3 for x in the first equation results in:
10/3 + y = 4
Clear fractions by multiplying all three terms by 3:
10 + 3y = 12
Then 3y = 2, and y = 2/3.
The solution is (10/3, 2/3)
The answer is: A - f(x) = 1/2 cos (x)
divide total bags by number of hours
30 bags / 3 hors = 10 bags per hour
he racked 10 bags per hour
Answer: 14x^2-93xy+60y^2 Hope that helps!
Step-by-step explanation:
1. Expand by distributing terms
(20x-12y)(x-4y)-(3x-4y)(2x+3y)
2. Use the Foil method:(a+b)(c+d)= ac+ad+bc+bd
20x^2-80xy-12yx+48y^2-(3x-4y)(2x+3y)
3. Use the Foil method : (a+b)(c+d)= ac+ad+bc+bd
20x^2-80xy-12yx+48y^2-(6x^2+9xy-8yx-12y^2)
4. Remove parentheses 20x^2-80xy-12yx+48y^2-6x^2-9xy+ 8yx+12y^2
5. Collect like terms (20x^2-6x^2)+(-80xy-12xy-9xy+8xy)+(48y^2+12y^2)
6. Simplify.
And your answer would be 14x^2-93xy+60y^2
Answer: 620≥ 427.57+ 11.38(4)+ (31.41) +57.75x
Step-by-step explanation:
So our limit is the 620$ that he has to spend. We know for a fact that the bike costs about 427.57 dollars. 4 $11.38 reflectors are purchased as well as gloves for 31.41. Now however many outfits he can purchase cannot exceed or go over his money limit of 620 so the variable (x) represents how many outfits he can buy without going over.
