The angle iron pictured here is made of steel with a density of 7.78g/cm^3. (Figure 1)

measurements are:
12.78cm
10.26cm
1.35cm
2.75cm
1.35cm

Without a picture I don't know. What's the question?

you definitely r taking Prof. Kowach lol

Ha! I just saw your comment... I am indeed are you in his class or you've taken him before, Donovan?

To find the volume of the angle iron, you need to calculate the product of its length, height, and thickness. Since the angle iron is a three-dimensional shape, you can use the formula V = l * h * t, where V represents the volume, l is the length, h is the height, and t is the thickness.

Using the given measurements:
Length (l) = 12.78 cm
Height (h) = 10.26 cm
Thickness (t) = 1.35 cm

Now, substitute these values into the formula:
V = 12.78 cm * 10.26 cm * 1.35 cm

Multiplying these values:
V = 175.22692 cm^3

The volume of the angle iron is 175.22692 cm^3.

To determine the mass of the angle iron, you can multiply the volume by the density of steel. The formula to calculate the mass (m) is m = V * density.

Given:
Density = 7.78 g/cm^3
Volume (V) = 175.22692 cm^3

Now, substitute these values into the formula:
m = 175.22692 cm^3 * 7.78 g/cm^3

Multiplying these values:
m = 1360.9923756 g

The mass of the angle iron is approximately 1360.99 grams (or 1.36 kg, rounding to two decimal places).

Note: Remember to include proper unit conversions and round your final answer to the appropriate number of decimal places.