A box's volume is equal to 600 cubic
1 answer:
Answer:
Step-by-step explanation:
Length = x
width = x - 7
height = x - 10
volume = 600 in^3
Volume = L * w * h
600 = x * (x - 7)*(x - 10)
600 = x * (x^2 - 17x + 70)
600 = x^3 - 17x^2 + 70x
x^3 - 17x^2 + 70x - 600 = 0
There is a genius somewhere that could solve this using algebra. The rest of us have 2 choices. We can graph it (see below) or we could use a program that solves it for us. I'll do both for you.
a = 1
b = - 17
c = 70
d = - 600
There is only 1 real solution: x = 15
L = 15
w = 8
h = 5
That has to be one of the ugliest graphs made. It does into the hundreds of thousands and it still looks like a straight line. Trust me. It's not.
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We can see that revolving the region formed by intersecting 3 lines, we will get 2 cones that are connected their bases.
Volume of the cone V=1/3 *πr²*h
1) small cone has r=5, and h=5
Volume small cone V1= 1/3 *π*5²*5 = 5³/3 *π
2) large cone has r=5, and h=21-6=15, h=15
Volume large cone V2= 1/3 *π*5²*15 = 5³*π
3) whole volume
5³/3 *π + 5³*π=5³π(1/3+1)=((5³*4)/3)π=(500/3)π≈166.7π≈523.6
Area
we see 2 right triangles,
Area of the triangle=1/2*b*h, where b -base, h -height
1) small one, b=5, h=5
A1=(1/2)*5*5=25/2
2)large one, b=5, h=15
A2=(1/2)*5*15=75/2
3) whole area=A1+A2=25/2+75/2=100/2=50
Volume of the cone V=1/3 *πr²*h
1) small cone has r=5, and h=5
Volume small cone V1= 1/3 *π*5²*5 = 5³/3 *π
2) large cone has r=5, and h=21-6=15, h=15
Volume large cone V2= 1/3 *π*5²*15 = 5³*π
3) whole volume
5³/3 *π + 5³*π=5³π(1/3+1)=((5³*4)/3)π=(500/3)π≈166.7π≈523.6
Area
we see 2 right triangles,
Area of the triangle=1/2*b*h, where b -base, h -height
1) small one, b=5, h=5
A1=(1/2)*5*5=25/2
2)large one, b=5, h=15
A2=(1/2)*5*15=75/2
3) whole area=A1+A2=25/2+75/2=100/2=50