The polynomial x 3 + 5x 2 - 57x -189 expresses the volume, in cubic inches, of a shipping box, and the width is (x+3) in. If t
he width of the box is 15 in., what are the other two dimensions? ( Hint: The height is greater than the depth.)
A. height: 19 in.
depth: 7 in.
B. height: 19 in.
depth: 5 in
C. height: 21 in.
depth: 5 in.
D. height: 21 in.
depth: 7 in.
A. height: 19 in.
depth: 7 in.
B. height: 19 in.
depth: 5 in
C. height: 21 in.
depth: 5 in.
D. height: 21 in.
depth: 7 in.
2 answers:
V=x^3+5x^2-57x-189
Width: W=(x+3) in = 15 in →x+3=15
Solving for x:
x+3-3=15-3→x=12
With x=12 the Volume would be:
V=(12)^3+5(12)^2-57(12)-189
V=1,728+5(144)-684-189
V=1,728+720-684-189
V=1,575
V=W*D*H
Depth: D
Height: H
with H>D
V=1,575; W=15
Replacing in the equation above:
1,575=15*D*H
Dividing both sides by 15
1,575/15=(12*D*H)/15
105=D*H
3*5*7=D*H
D<H
If D=5→H=3*7→H=21
If D=7→H=3*5→H=15
Answer: Option <span>C. height: 21 in. depth: 5 in.
</span>
Please, see the attached file for another form to solve the problem
Width: W=(x+3) in = 15 in →x+3=15
Solving for x:
x+3-3=15-3→x=12
With x=12 the Volume would be:
V=(12)^3+5(12)^2-57(12)-189
V=1,728+5(144)-684-189
V=1,728+720-684-189
V=1,575
V=W*D*H
Depth: D
Height: H
with H>D
V=1,575; W=15
Replacing in the equation above:
1,575=15*D*H
Dividing both sides by 15
1,575/15=(12*D*H)/15
105=D*H
3*5*7=D*H
D<H
If D=5→H=3*7→H=21
If D=7→H=3*5→H=15
Answer: Option <span>C. height: 21 in. depth: 5 in.
</span>
Please, see the attached file for another form to solve the problem
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