The balls are stacked directly on top of one another, so the limiting measurement is their combined height, not their volume.

Each ball has a diameter of \(2 \times 8 = 16\) cm.

Stacked height of \(6\) balls: \(6 \times 16 = 96\) cm

The container height is \(71\) cm, which is less than \(96\) cm, so the balls do not fit inside the container.

No

Substitute \(r = 5\) into the formula you have been given. Square the radius before multiplying by \(4\):

\(A = 4 \times \pi \times 5^2 = 4 \times \pi \times 25 = 100\pi\)

\(100\pi\) cm²

The correct method uses the radius, cubed, times the hemisphere fraction \(\frac{2}{3}\):

\(V = \frac{2}{3} \times \pi \times 15^{3} = 2250\pi\) cm³

Checking the shown working against that, line by line:

First mistake: used the diameter in place of the radius.

Second mistake: squared instead of cubing (used power 2, not power 3).

The formula uses the radius, and the sphere has been measured right across, so halve the diameter first:

\(r = 20 \div 2 = 10\) cm

Now cube the radius, then multiply:

\(r^3 = 10^3 = 1000\)

\(V = \frac{4}{3} \times \pi \times 1000\)

Using the \(\pi\) key on a calculator, \(V = 4188.8\) cm³ to 1 decimal place.

The whole ball first:

\(V = \frac{4}{3} \times \pi \times 12^3 = \frac{4}{3} \times \pi \times 1728 = 2304\pi\) cm³

A hemisphere is exactly half of that:

\(2304\pi \div 2 = 1152\pi\)

\(1152\pi\) cm³

Half the *curved* surface of the ball is only part of the answer. Cutting the ball open makes a new flat face — a circle — and that circle counts too.

The curved part is half the sphere's surface:

\(\frac{1}{2} \times 4\pi r^2 = 2\pi \times 144 = 288\pi\) cm²

The flat face is a circle of radius \(12\) cm:

\(\pi r^2 = 144\pi\) cm²

\(288\pi + 144\pi = 432\pi\)

\(432\pi\) cm²

Each of the cube's 6 faces has area \(x^2\), so its total surface area is \(6x^2\).

Substitute \(r = 3\) into the formula for the sphere:

\(4\pi r^2 = 4\pi \times 3^2 = 36\pi\)

The two surface areas are equal:

\(6x^2 = 36\pi\)

Divide both sides by 6:

\(x^2 = 6\pi\)

Take the square root of both sides:

\(x = \sqrt{ 6\pi }\).

First find the full volume of the ball, using the radius.

\(V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \pi \times 22^3 \approx 44602.2\) cm³

Convert to litres: \(44602.2 \div 1000 \approx 44.6\) litres

The ball is not being filled from empty — it already has \(15.6\) litres of air inside, so this must be subtracted before dividing by the rate:

Volume still needed: \(44.6 - 15.6 \approx 29.0\) litres

Extra time to fill: \(29.0 \div 1.5 \approx 19.3\) minutes

\(19.3\) minutes is more than \(15\) minutes, so the answer is No.

First find the volume of water poured in. It is poured at \(46.4\) cm³ per second for \(18\) seconds:

\(46.4 \times 18 \approx 835.2\) cm³

Next find the full capacity of the bowl. A hemisphere is half of a sphere of the same radius:

\(\frac{1}{2} \times \frac{4}{3} \times \pi \times 10^3 \approx 2094.4\) cm³

Now write the water poured as a percentage of that capacity:

\(835.2 \div 2094.4 \times 100 \approx 39.9\%\)

\(39.9\%\) is below \(45\%\), so the water does not fill more than \(45\%\) of the bowl.

No

The bases are stuck together, so they are inside the solid. Only the two curved surfaces are on the outside.

First cone: \(\pi \times r \times 3r = 3\pi r^2\)

Second cone: \(\pi \times r \times 4r = 4\pi r^2\)

Total: \(3\pi r^2 + 4\pi r^2 = 7\pi r^2\)

This equals \(323.68\pi\), so \(7\pi r^2 = 323.68\pi\)

Divide both sides by \(\pi\): \(7r^2 = 323.68\)

\(r^2 = 323.68 \div 7 = 46.24\)

\(r = \sqrt{ 46.24 } = 6.8\)

The radius is \(6.8\) cm.

The sphere has radius \(8\) cm:

\(\frac{4}{3} \times \pi \times 8^3 = 682\frac{2}{3}\pi\) cm³

The cone has base radius \(8\) cm and perpendicular height \(29\) cm:

\(\frac{1}{3} \times \pi \times 8^2 \times 29 = 618\frac{2}{3}\pi\) cm³

Now compare the multiples of \(\pi\). The sphere gives \(682\frac{2}{3}\) and the cone gives \(618\frac{2}{3}\), so the sphere's volume is greater than the cone's.

The sphere holds more.