The box will have six total faces: an identical "top and bottom," and identical "left and right," and an identical "front and back." The total surface area will be the sum of these faces.

Since the six faces consider of three sets of pairs, we can set up the equation as:

\(\displaystyle SA=2(\text{top})+2(\text{left})+2(\text{front})\)

Each of these faces will correspond to one pair of dimensions. Multiply the pair to get the area of the face.

\(\displaystyle SA=2(lw)+2(wh)+2(lh)\)

Substitute the values from the question to solve.

\(\displaystyle SA=2(5*5)+2(5*4)+2(5*4)\)

\(\displaystyle SA=50+40+40\)

\(\displaystyle SA=130\ in^2\)

The formula for the surface area of a rectangular prism is given by:

SA = 2LW + 2WH + 2HL

SA = 2(12 * 8) + 2(8 * 6) + 2(6 * 12)

SA = 2(96) + 2(48) + 2(72)

SA = 192 + 96 + 144

SA = 432 in²

216 in²  is the wrong answer because it is off by a factor of 2

576 in³ is actually the volume, V = L * W * H 

The formula for the volume of a rectangular prism is \(\displaystyle V = L*W*H\), where "\(\displaystyle V\)" is volume, "\(\displaystyle L\)" is length, "\(\displaystyle W\)" is width and "\(\displaystyle H\)" is height.

We know that \(\displaystyle L = 2W\) and \(\displaystyle H = L+W\). By rearranging \(\displaystyle L=2W\), we get \(\displaystyle W=\frac{L}{2}\). Substituting \(\displaystyle \frac{L}{2}\) into the volume equation for \(\displaystyle W\) and \(\displaystyle L+W\) into the same equation for \(\displaystyle H\), we get the following:

\(\displaystyle V = L \cdot \left(\frac{L}{2}\right)\cdot (L+\left(\frac{L}{2}\right))\)

\(\displaystyle V = L\cdot \left(\frac{L}{2}\right)\cdot \frac{3L}{2}\)

\(\displaystyle V = 10 \cdot 5 \cdot 15\)

\(\displaystyle V = 750\) units cubed

Given that the dimensions are: \(\displaystyle 6.3 \:cm\), \(\displaystyle 6.3 \:cm\), and \(\displaystyle 2.1 \:cm\) and that the volume of a rectangular prism can be given by the equation:

\(\displaystyle v=l \cdot w \cdot h\), where \(\displaystyle l\) is length, \(\displaystyle w\) is width, and \(\displaystyle h\) is height, the volume can be simply solved for by substituting in the values.

\(\displaystyle V = (6.3)(6.3)(2.1)\)

\(\displaystyle V = (39.69)(2.1)\)

\(\displaystyle V = 83.349\:cm^3\)

This final value can be approximated to \(\displaystyle 83.3\:cm^3\).

The volume of a rectangular prism is found using the following formula:

\(\displaystyle V=l\times w\times h\)

If we substitute our known values, then we can solve for the missing side.

\(\displaystyle 84=3\times 4\times x\)

\(\displaystyle 84=12x\)

Divide both sides of the equation by 12.

\(\displaystyle \frac{84}{12}=\frac{12x}{12}\)

\(\displaystyle x=7\)

We now know that the missing length equals 7 centimeters.

This means that the box can have sides with the following dimensions: 3cm by 4cm; 7cm by 3cm; or 7cm by 4cm. The greatest area of one side belongs to the one that is 7cm by 4cm. 

\(\displaystyle A=l\times w\)

\(\displaystyle A=4\times 7\)

\(\displaystyle A=28\)

The volume of the rectangle 

\(\displaystyle V = l*w*h\) 

so we plug in our values and obtain

\(\displaystyle V = 4*3*8\)

\(\displaystyle V = 96 m^3\).

A regular tetrahedron has 4 triangular faces. The area of one of these faces is given by:

\(\displaystyle \small A=\frac{1}{2}bh\)

Because the surface area is the area of all 4 faces combined, in order to find the area for one of the faces only, we must divide the surface area by 4. We know that the surface area is \(\displaystyle \small 684cm^2\), therefore:

\(\displaystyle \small A=\frac{S.A.}{4}\)

\(\displaystyle A=\frac{684cm^{2}}{4}=171cm^2\)

Since we now have the area of one face, and we know the height of one face is \(\displaystyle 18cm\), we can now plug these values into the original formula:

\(\displaystyle \small A=\frac{1}{2}bh\)

\(\displaystyle \small 171=\frac{1}{2}b(18)\)

\(\displaystyle \small 171=9b\)

\(\displaystyle \small b=19\)

Therefore, the length of the base of one face is \(\displaystyle \small 19cm\).

The only given information is the surface area of the regular tetrahedron.

This is a quick problem that can be easily solved for by using the formula for the surface area of a tetrahedron:

\(\displaystyle SA= \sqrt{3}\cdot a^2\)

If we substitute in the given infomation, we are left with the edge being the only unknown. 

\(\displaystyle 156 = \sqrt{3} \cdot a^2\)

\(\displaystyle \frac{156}{\sqrt{3}}=a^2\)

\(\displaystyle \sqrt{\frac{156}{\sqrt{3}}}=a\)

\(\displaystyle a=9.49 \approx{\color{Blue} 9.5}\)

The problem provides the information for the slant height and the area of one of the equilateral triangle faces. 

The slant height merely refers to the height of this equilateral triangle. 

Therefore, if we're given the area of a triangle and it's height, we should be able to solve for it's base. The base in this case will equate to the measurement of the edge. It's helpful to remember that in this case, because all faces are equilateral triangles, the measure of one length will equate to the length of all other edges.

We can use the equation that will allow us to solve for the area of a triangle:

\(\displaystyle A=\frac{1}{2} \cdot b\cdot h\)

where \(\displaystyle b\) is base length and \(\displaystyle h\) is height.

Substituting in the information that's been provided, we get:

\(\displaystyle 43.3 = \frac{1}{2} \cdot b \cdot \frac{10\sqrt{3}}{2}\)

\(\displaystyle b \cdot \frac{10\sqrt{3}}{2}= 2 \cdot 43.3\)

\(\displaystyle b = 2 \cdot 43.3 \cdot \frac{2}{10\sqrt{3}}\)

\(\displaystyle b =9.99971\approx {\color{Blue} 10}\)

This becomes a quick problem by just utilizing the formula for the volume of a tetrahedron. 

\(\displaystyle V= \frac{a^3}{6\sqrt{2}}\)

Upon substituting the value for the volume into the formula, we are left with \(\displaystyle a\), which represents the edge length. 

\(\displaystyle 94.8= \frac{a^3}{6\sqrt{2}}\)

\(\displaystyle a^3=94.8 \cdot 6\sqrt{2}\)

\(\displaystyle a= \sqrt[3]{94.8 \cdot 6\sqrt{2}}\)

\(\displaystyle a= {\color{Blue} 9.3}\)