The Volume of the box has to equal 64
because all the sides have to equal 12
and 12/3 = 4
Step-by-step explanation:
singkamas: 60 pesos
ponkan: 110 pesos
Step-by-step explanation:
y = 3 + 8x^(³/₂), 0 ≤ x ≤ 1
dy/dx = 12√x
Arc length is:
s = ∫ ds
s = ∫₀¹ √(1 + (dy/dx)²) dx
s = ∫₀¹ √(1 + (12√x)²) dx
s = ∫₀¹ √(1 + 144x) dx
If u = 1 + 144x, then du = 144 dx.
s = 1/144 ∫ √u du
s = 1/144 (⅔ u^(³/₂))
s = 1/216 u^(³/₂)
Substitute back:
s = 1/216 (1 + 144x)^(³/₂)
Evaluate between x=0 and x=1.
s = [1/216 (1 + 144)^(³/₂)] − [1/216 (1 + 0)^(³/₂)]
s = 1/216 (145)^(³/₂) − 1/216
s = (145√145 − 1) / 216
The pile contains 17 quarters and 15 half-dollars.
Let <em>x</em> = the number of quarters and <em>y</em> = the number of half-dollars.
We have two equations:
(1) $0.25<em>x</em> + $0.50<em>y</em> = $11.75
(2) <em>x</em> = <em>y</em> +2
Substitute the value of <em>x</em> from Equation (2) into Equation (1).
0.25(<em>y</em>+2) + 0.50<em>y</em> = 11.75
0.25<em>y</em> + 0.50 + 0.50<em>y</em> = 11.75
0.75<em>y</em> = 11.75 – 0.50 = 11.25
<em>y</em> = 11.25/0.75 = 15
Substitute the value of <em>y</em> in Equation (2).
<em>x</em> = 15 + 2 = 17
The pile contains 17 quarters and 15 half-dollars.
<em>Check</em>: 17×$0.25 + 15×$0.50 = $4.25 + $7.50 = $11.75.
