Answer: a
Step-by-step explanation:
Answer:
the answer is 8/17
Step-by-step explanation:
i got it right on the quiz
Let the three gp be a, ar and ar^2
a + ar + ar^2 = 21 => a(1 + r + r^2) = 21 . . . (1)
a^2 + a^2r^2 + a^2r^4 = 189 => a^2(1 + r^2 + r^4) = 189 . . . (2)
squaring (1) gives
a^2(1 + r + r^2)^2 = 441 . . . (3)
(3) ÷ (2) => (1 + r + r^2)^2 / (1 + r^2 + r^4) = 441/189 = 7/3
3(1 + r + r^2)^2 = 7(1 + r^2 + r^4)
3(r^4 + 2r^3 + 3r^2 + 2r + 1) = 7(1 + r^2 + r^4)
3r^4 + 6r^3 + 9r^2 + 6r + 3 = 7 + 7r^2 + 7r^4
4r^4 - 6r^3 - 2r^2 - 6r + 4 = 0
r = 1/2 or r = 2
From (1), a = 21/(1 + r + r^2)
When r = 2:
a = 21/(1 + 2 + 4) = 21/7 = 3
Therefore, the numbers are 3, 6 and 12.
Answer:
(a)
(b)
(c)
Step-by-step explanation:
We are required to construct 3 linear equations starting with the given solution z = 1/3.
<u>Equation 1</u>
<u />
<u />
Multiply both sides by 9

Rewrite 3 as 5-2
9z=5-2
Add 2 to both sides
Our first equation is: 
<u>Equation 2</u>
<u />
<u />
Multiply both sides by 21

Rewrite 7 as 11-4
21z=11-4
Subtract 11 from both sides
Our second equation is: 
<u>Equation 3</u>
<u />
<u />
Multiply both sides by 6

Rewrite 6z as 4z+2z
4z+2z=2
Subtract 2z from both sides
Our third equation is: 