Answer:
Your answer is the first point
A reflection across the x-axis and then a transformation of -6 and 1.
Given data :
a₃ = 9/16
aₓ = -3/4 · aₓ₋₁
Where x is the number of terms ('x' is also written as 'n')
To find the 7th term (a₇):
We know that aₓ = -3/4 · aₓ₋₁
So,
a₃ = -3/4 · a₃₋₁
a₃ = -3/4 · a₂
9/16 = -3/4 · a₂
a₂ = 9/16 × -4/3
a₂ = -36/48
a₂ = -3/4
Again,
aₓ = -3/4 · aₓ₋₁
a₄ = -3/4 · a₄₋₁
a₄ = -3/4 · a₃
a₄ = -3/4 · 9/16
a₄ = -27/64
a₄ = -27/64
For a₅,
aₓ = -3/4 · aₓ₋₁
a₅ = -3/4 · a₅₋₁
a₅ = -3/4 · a₄
a₅ = -3/4 × -27/64
a₅ = 81/256
For a₆,
aₓ = -3/4 · aₓ₋₁
a₆ = -3/4 · a₆₋₁
a₆ = -3/4 · a₅
a₆ = -3/4 × 81/256
a₆ = -243/1024
For a₇,
aₓ = -3/4 · aₓ₋₁
a₇ = -3/4 · a₇₋₁
a₇ = -3/4 · a₆
a₇ = -3/4 × -243/1024
a₇ = 729/4096
Slope = y2 - y1/x2 - x1
6 - -3/1 - -6
=9/7
D/2=r
vcone=(1/3)hpir^2
given
d=30
h=9.1
d/2=30/2=15=r
v=(1/3)9.1pi15^2
v=(9.1/3)pi225
v=682.5pi
use 3.141592 to aprox pi
v=2144.13654
the closese is the first one
answer would be 2120 m³
Answer:
<em>96π units²</em>
Step-by-step explanation:
Find the diagram attached
Area of a sector is expressed as;
Area of a sector = θ/2π * πr²
Given
θ = 3π/4
r = 16
Substitute into the formula
area of the sector = (3π/4)/2π * π(16)²
area of the sector = 3π/8π * 256π
area of the sector = 3/8 * 256π
area of the sector = 3 * 32π
<em>area of the sector =96π units²</em>
