Answer:
see explanation
Step-by-step explanation:
(a)
Given
2k - 6k² + 4k³ ← factor out 2k from each term
= 2k(1 - 3k + 2k²)
To factor the quadratic
Consider the factors of the product of the constant term ( 1) and the coefficient of the k² term (+ 2) which sum to give the coefficient of the k- term (- 3)
The factors are - 1 and - 2
Use these factors to split the k- term
1 - k - 2k + 2k² ( factor the first/second and third/fourth terms )
1(1 - k) - 2k(1 - k) ← factor out (1 - k) from each term
= (1 - k)(1 - 2k)
1 - 3k + 2k² = (1 - k)(1 - 2k) and
2k - 6k² + 4k³ = 2k(1 - k)(1 - 2k)
(b)
Given
2ax - 4ay + 3bx - 6by ( factor the first/second and third/fourth terms )
= 2a(x - 2y) + 3b(x - 2y) ← factor out (x - 2y) from each term
= (x - 2y)(2a + 3b)
The complement is 38
and the supplement is 128
Answer:
A=z
B=-2
C=0
D=30
Step-by-step explanation:
First lets simply the equation
21 - (2x +3)2 + 4(1-1) + z^3
21 - (2x^2 + 9) + 4(0) + z ^3
21 - 2x^2 + 9 + z^3
- 2x^2 + z ^3 + 30
This means that:
A=z
B=-2
C=0
D=30
Answer:
1 + ln 2 - ln x
Step-by-step explanation:
ln ( 2e /x)
We know that ln ( a/b) = ln ( a) - ln (b)
ln (2e) - ln (x)
We also know that ln ( a*b) = ln a + ln b
ln ( 2) + ln e - ln x
We know that the ln e = 1
ln 2 + 1 - ln x
Changing the order
1 + ln 2 - ln x
