Three-eighths of the sum of sixteen times a number and twenty-four. How do you set this up?
1 answer:
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Weird. A period appears above this... huh.
Answer:
[Th]e cube has a greater value.
Step-by-step explanation:
What the word problem really wants us to get [is ]the question of 'Which is greater, A=6a^2 when 'a' [is] 18 or A=4r^2 when r = 9? And here's how to solve that.
Starting with the[ c]ube we have A=6a^2. A bit t[o]o simple, right?
A=6(18)^2 Substitute numbers.
A=6(324) Solve ex[p]onents.
A=1944 Mult[i]ply.
So w[e] know that the cube is 1944 meters cube[d ] in area. But what about the more [f]ormidable sphere? Fo[r] it we need a slightly m[o]re co[m]plicated formula, A=4r^2. However, instead of using the real pi I will be rounding to 3.14, since we have no calculator so anything more would take way too long and fry your[ bra]in.
A=4(3.14)(9)^2 Subst[i]tute numbers.
A=4(3.14)(81) Solve expone[n]ts.
A=12.56(81) Multip[ly].
A=1017.36 Multiply again[.]
Now, since I'm sure all of us can count, we know that 1944 is greater than 1017.36. Or in other words, the cube is bigger than the sphere.
And PLEASE don't copy this guys. Make your own iteration. Change it up!