16 soccer cards and 22 baseball cards into group. What is the greatest number of cards she can put in each group? How many group
1 answer:
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Answer:
S₁₅ = 645
Step-by-step explanation:
The sum to n terms of an arithmetic sequence is
=
[ 2a₁ + (n - 1)d ]
where a₁ is the first term and d the common difference
Using the nth term formula = 5n + 3 , then
a₁ = 5(1) + 3 = 5 + 3 = 8
a₂ = 5(2) + 3 = 10 + 3 = 13 , then
d = a₂ - a₁ = 13 - 8 = 5
Thus
S₁₅ = [ (2 × 8) + (14 × 5) ]
= 7.5( 16 + 70)
= 7.5 × 86
= 645
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!