Answer:
you are looking for square inches of wrapping paper, you are looking for the surface area of the box. A three-dimensional box has a top and bottom that are the same size, a front and back that are the same size, and 2 sides that are the same size...or SA = 2(top/bottom area) + 2(front/back area) + 2(side/side area). Our equation fills in nicely like this:
SA = 2(6 * 14) + 2(6 * 8) + 2(14 * 8)
SA = 2(84) + 2(48) + 2(112)
SA = 168 + 96 + 224
SA = 488 square inches of paper to wrap the box.
hope this helps :]
the answer is b Im pretty sure
Answer:
a) possible progressions are 5
b) the smallest and largest possible values of the first term are 16 and 82
Step-by-step explanation:
<u>Sum of terms:</u>
- Sₙ = n/2(a₁ + aₙ) = n/2(2a₁ + (n-1)d)
- S₂₀ = 20/2(2a₁ + 19d) = 10(2a₁ + 19d)
- 2020 = 10(2a₁ + 19d)
- 202 = 2a₁ + 19d
<u>In order a₁ to be an integer, d must be even number, so d = 2k</u>
- 202 = 2a₁ + 38k
- 101 = a₁ + 19k
<u>Possible values of k= 1,2,3,4,5</u>
- k = 1 ⇒ a₁ = 101 - 19 = 82
- k = 2 ⇒ a₁ = 101 - 38 = 63
- k = 3 ⇒ a₁ = 101 - 57 = 44
- k = 4 ⇒ a₁ = 101 - 76 = 25
- k = 5 ⇒ a₁ = 101 - 95 = 16
<u>As per above, </u>
- a) possible progressions are 5
- b) the smallest and largest possible values of the first term are 16 and 82
Answer:
i do not know what to find
Step-by-step explanation:
