I wasn't going to click on this one, but the all-caps enthralled me and hypnotized me.
first off, let's change all the mixed fractions to "improper" and proceed from there, keep in mind that if we subtract the two bags' weight from the total, what's leftover is the 3rd bag's weight.
Answer:
Perimeter of the figure = 24 m.
Step-by-step explanation:
The given figure is a square of side 6 m in which a rectangle of length 4 m and width 2 m is cut at the corner.
Therefore, the missing sides are 6-4 = 2 m and 6-2 = 4 m.
Hence, the perimeter of the figure = 6 + 6 + 2 + 2 + 4 + 4 = 24 m.
<h3>
Answer: Yes they are equivalent</h3>
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Work Shown:
Expand out the first expression to get
(a-3)(2a^2 + 3a + 3)
a(2a^2 + 3a + 3) - 3(2a^2 + 3a + 3)
2a^3 + 3a^2 + 3a - 6a^2 - 9a - 9
2a^3 + (3a^2-6a^2) + (3a-9a) - 9
2a^3 - 3a^2 - 6a - 9
Divide every term by 2 so we can pull out a 2 through the distributive property
2a^3 - 3a^2 - 6a - 9 = 2(a^3 - 1.5a^2 - 3a - 4.5)
This shows that (a-3)(2a^2 + 3a + 3) is equivalent to 2(a^3 - 1.5a^2 - 3a - 4.5)
Answer:
You simply substitute 8 for n
A₈ =3(8)+4=28
