Let the cost of 1 notebook be x and the cost of 1 binder be y.
4 notebooks and 3 binders would cost 23.5
Therefore, 4x + 3y = 23.5 (1)
7 notebooks and 6 binders would cost 44.5
Therefore, 7x + 6y = 44.5 (2)
Multiply the first equation by 2.
8x + 6y = 47 (3)
(3) - (2) gives
x = 2.5
Substitute the value of x in (1), we get,
4(2.5) + 3y = 23.5
10 + 3y = 23.5
3y = 23.5 - 10
3y = 13.5
y = 13.5/3
y = 4.5
Hence, cost of 5 notebooks and 3 binders is:
5x + 3y = 5(2.5) + 3(4.5)
= 12.5 + 13.5
= 26
Hence, cost of 5 notebooks and 3 binders is $26.
You do distance over time to get the rate, so you'd do 27/2= 13.5mph then divide that in half since you're asking for the distance traveled in half that hour so you do 13.5/2 and you get 6.75 miles traveled in half and hour.
A = L * W
A = 320
L = W + 4
320 = W(W + 4)
320 = W^2 + 4W
W^2 + 4W = 320
W^2 + 4W + 4 = 320 + 4
(W + 2)^2 = 324
W + 2 = (+-) sqrt 324
W = -2 (+-) 18
W = -2 + 18 = 16 ft <== this is the width
W = -2 - 18 = -20....not this one because it is negative
L = W + 4
L = 16 + 4
L = 20 ft <=== this is the length
in summary...the width is 16 ft and the length is 20 ft
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.
