Respuesta :
Answer:
[tex]w = 9 \text{ inches}\\l = 13\text{ inches}\\h = 3\text{ inches}[/tex]
Step-by-step explanation:
We are given the following in the question:
Volume of cake = 351 cubic inches
Let x inches be the width of cake.
Width of cake, w =
[tex]x\text{ inches}[/tex]
Then, length of cake,l =
[tex](x + 4)\text{ inches}[/tex]
Height of cake,h =
[tex]\dfrac{x}{3}\text{ inches}[/tex]
Volume of cake = Volume of cuboid
[tex]V = lwh[/tex]
Putting values, we get:
[tex]351 = x(x+4)(\dfrac{x}{3})\\\\1053 = x^3 + 4x\\x^3+4x^2-1053= 0\\\\\text{For x = 9}\\(9)^3+4(9)^2-1053= 0[/tex]
Thus, dimensions of cake are:
[tex]w = 9 \text{ inches}\\l = 13\text{ inches}\\h = 3\text{ inches}[/tex]
The volume of the cake and the relationships between the dimensions of
the cake can be used to find the length, width, and height of the cake.
The dimensions of the cake pan should be;
- Length; 13 inches
- Width; 9 inches
- Height; 3 inches
Reasons:
The volume of the cake = 351 in.³
Length of cake, L = 4 inches longer than the width
Height of the cake = Width of the cake ÷ 3
Let L represent the length of the cake, let W represent the width of the
cake, and let H, represent the height of the cake.
Therefore;
L = 4 + W
H = [tex]\dfrac{W}{3}[/tex]
The volume of the cake, V, is therefore;
[tex]V = 351 = L \times W \times H = (W + 4) \times W \times \dfrac{W}{3} = \dfrac{W^3}{3} + \dfrac{4 \cdot W^2}{3}[/tex]
W³ + 4·W² = 351 × 3
W³ + 4·W² - 351 × 3 = W³ + 4·W² - 1053 = 0
W³ + 4·W² - 1053 = (W - 9)·(W² + 13·W + 117) = 0
Therefore;
The width of the cake, W = 9 inches
L = W + 4
Therefore;
The length of the cake, L = 9 inches + 4 inches = 13 inches
H = [tex]\dfrac{W}{3}[/tex]
The height of the cake, H = [tex]\dfrac{9 \ inches}{3}[/tex] = 3 inches
The dimensions of the cake pan are;
- Length; 13 inches
- Width; 9 inches
- Height; 3 inches
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