70+75+80+90+100=415
415/5=83
<span>After the lowest grade is dropped, Janet's average would be an 83%</span>
X-y=3=>x=3+y
X+2y=-6
=========
3+y+2y=-6
3+3y=-6
3y=-6-3
3y=-9
y=-3
X+2y=-6
X+2(-3)=-6
X-6=-6
X=0
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.
The interval that f(x) is increasing is the distance from 200 to 300.
The minimum value of f(x) in the interval 0<x<300 is 200.
At a value of 500, the value of f(x) is 0.
The function can't be a quadratic function since there are two points in the graph where f(x) changes its rate from increasing to decreasing or the opposite. A quadratic function has only one of that point.
Answer:
A-2515.57647469
Step-by-step explanation:
