You plug in the x’s to the equation up top to gin the y
Answer:
maybe 3...............................
Answer:
133 fishes
Step-by-step explanation:
Units of food A = 400 units
Units of food B = 400 units
Fish Bass required 2 units of A and 4 units of B.
Fish Trout requires 5 units of A and 2 units of B.
i. For food A,
total units of food A required = 2 + 5
= 7 units
number of bass and trout that would consume food A = 2 x 
= 114.3
number of bass and trout that would consume food A = 114
ii. For food B,
total units of food B required = 4 + 2
= 6 units
number of bass and trout that would consume food B = 2 x 
= 133.3
number of bass and trout that would consume food B = 133
Thus, the maximum number of fish that the lake can support is 133.
8+11=19 but if you add 21+19+12+5=57.
Answer:
A) a>1
Step-by-step explanation:
2(a +8) >18
2a +16 >18
2a>18 - 16
2a>2
a>2:2
a>1