Answer:
1st and 3rd go in the 1st box
2nd and 4th go in the 2nd box
last one goes in the last box
Step-by-step explanation:
1,2,3,4,5 and time spent jogging go in independent quantity
3.5, 7, 10.5, 14, 17.5 and distance traveled go in dependent quantity
m=3.5h goes in neither
SA=2H(L+W)+LW for open box
20=2H(4+W)+4(W)
distribute
20=8H+2WH+4W
divide both sides by 2
10=4H+WH+2W
solve for 1 variable, pick W
10-4H=WH+2W
10-4H=W(H+2)
(10-4H)/(H+2)=W
V=LWH
subsitute 4 for V, subsitute H for H and (10-4H)/(H+2) for W
V=(4)(H)(10-4H)/(H+2)
V=(40H-16H²)/(H+2)
find max value
take deritivitive of this thing
V'=-16(H²+4H-5)/((H+2)²)
using sign chart
sign changes from positive to negative at H=1
so at H=1
find W
W=(10-4H)/(H+2)
W=2
the dimeionts are
length=4ft
width=2ft
height=1ft
(the volume is 8 cubic feet)
Answer:
36 pencils
Step-by-step explanation:
Let h and p represent the number of highlighters and the number of pencils, respectively.
Then h + p = 45, and h = 45 - p.
Tom paid a total of $30 for these supplies, with ($2/highligher)(h) + ($0.333/pencil) adding up to that amount.
substituting 45 - p for h in 2h + 0.333p = 30, we get:
2(45 - p) + 0.333p = 30, or
90 - 2p + 0.333p = 30
Combine the constants: 60 = 2p - 0.333p, or 60 = 1.667p
Then p = 60/1.667 = 35.9928, or 36.
Tom bought 36 pencils for $12, and 45-36, or 9, highlighters for $18, for a total purchase of $30. This shows that these calculations are correct.
If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero. If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity. If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity. The sum of the multiplicities is the degree
Answer:
well 126/9=14 but is the other stuff factor families or something? like whats the question asking
Step-by-step explanation:
