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svet-max [94.6K]
3 years ago
13

31. What is the maximum volume of a square pyramid that can pit inside a cube with a side length of 24cm?

Mathematics
1 answer:
REY [17]3 years ago
4 0
The answer is <span>B) 4,608cm^3

The volume of the square pyramid is:
V = a</span>² * h / 3
a - the side length of the base
h - the height of the pyramid.

Since it is inside the cube, then a = h = 24 cm.

So:
V = 24² * 24 / 3 = 13,824 / 3 = 4,608 cm³
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Volgvan

Answer:

\displaystyle y=\frac{16-9x^3}{2x^3 - 3}

\displaystyle y=-\frac{56}{13}

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}

i)

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3

Simplify the radical with the cube:

\displaystyle x^3=\frac{3y+16}{2y+9}

Multiply by 2y+9

\displaystyle x^3(2y+9)=\frac{3y+16}{2y+9}(2y+9)

Simplify:

\displaystyle x^3(2y+9)=3y+16

Operate the parentheses:

\displaystyle x^3(2y)+x^3(9)=3y+16

\displaystyle 2x^3y+9x^3=3y+16

Subtract 3y and 9x^3:

\displaystyle 2x^3y - 3y=16-9x^3

Factor y out of the left side:

\displaystyle y(2x^3 - 3)=16-9x^3

Divide by 2x^3 - 3:

\mathbf{\displaystyle y=\frac{16-9x^3}{2x^3 - 3}}

ii) To find y when x=2, substitute:

\displaystyle y=\frac{16-9\cdot 2^3}{2\cdot 2^3 - 3}

\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}

\displaystyle y=\frac{16-72}{16- 3}

\displaystyle y=\frac{-56}{13}

\mathbf{\displaystyle y=-\frac{56}{13}}

8 0
3 years ago
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