Answer:
t10 = 59x - 30
Step-by-step explanation:
a = 5x + 6
d = 11x + 2 - 5x - 6
d = 6x - 4
n = 10
tn = a + (n - 1)*d
t10 = 5x + 6 + (10 - 1) ( 6x - 4)
t10 = 5x + 6 + 9(6x - 4)
t10 = 5x + 6 + 54x - 36
t10 = 59x - 30
Check by finding the 3rd term
d = 6x - 4
n = 3
a = 5x + 6
t3 = 5x + 6 + (3-1)*(6x - 4)
t3 = 5x + 6 + 2(6x - 4)
t3 = 5x + 6 + 12x - 8
t3 = 17x - 2 which is exactly what it should be.
Answer:
F(a) = a² + 3a - 2
F(x-1) = x² + x -4
Step-by-step explanation:
Note:
Domain is missing, assume domain
F(a)
F(x-1)
Given:
F(x) = x² + 3x - 2
Computation:
1. F(x) = x² + 3x - 2
So,
⇒ F(a) = a² + 3a - 2
2. F(x-1) = (x-1)² + 3(x-1) - 2
F(x-1) = x² + 1 -2x + 3x - 3 - 2
F(x-1) = x² + x -4
Step-by-step explanation:
The solutions of the related quadratic equation are -1.5 and -3
Answer:
variable (a letter) <12
Step-by-step explanation:
:)
Step-by-step explanation:
There are 2 variables a and c, and there is only 1 equation. Hence we cannot find both variables.
However if we restrict a and c to be positive integers, we will have:
10a + 5.5c = 31
=> 20a + 11c = 62
Since 62 ≡ 2(mod 20) and 11c <= 62,
we have c = 2.
=> 20a + 11(2) = 62, 20a = 40, a = 2.
Therefore a = 2 and c = 2.
Hopefully this explanation helped!