Given:
bake sale at least $300
price of each pie is $6.00
let the number of pies be represented by x.
Write the inequality:
6x <u>></u> 300
Solve the inequality:
6x <u>></u> 300
<u>÷6 ÷6</u>
x <u>></u> 50
Graph the inequality.
y = 6x
x is the number of pies sold; x at least 50 and gradually increases.
y is the total sales
In this case just do it left to right, as multiplication comes before addition/subtraction. So 7x3x2+5-3=21x2+5-3=42+5-3=47-3=44=k.
Would this work? Hope it helps
Answer:
there is no greatest load
Step-by-step explanation:
Let x and y represent the load capacities of my truck and my neighbor's truck, respectively. We are given two relations:
x ≥ y +600 . . . . . my truck can carry at least 600 pounds more
x ≤ (1/3)(4y) . . . . . my truck carries no more than all 4 of hers
Combining these two inequalities, we have ...
4/3y ≥ x ≥ y +600
1/3y ≥ 600 . . . . . . . subtract y
y ≥ 1800 . . . . . . . . multiply by 3
My truck's capacity is greater than 1800 +600 = 2400 pounds. This is a lower limit. The question asks for an <em>upper limit</em>. The given conditions do not place any upper limit on truck capacity.
