Answer:
see explanation
Step-by-step explanation:
Given
4
- 5a² + 1 = 0
Use the substitution u = a², then equation is
4u² - 5u + 1 = 0
Consider the product of the coefficient of the u² term and the constant term
product = 4 × 1 = 4 and sum = - 5
The factors are - 4 and - 1
Use these factors to split the u- term
4u² - 4u - u + 1 = 0 ( factor the first/second and third/fourth terms )
4u(u - 1) - 1(u - 1) = 0 ← factor out (u - 1) from each term
(u - 1)(4u - 1) = 0
Equate each factor to zero and solve for u
u - 1 = 0 ⇒ u = 1
4u - 1 = 0 ⇒ 4u = 1 ⇒ u = 
Convert u back into terms of a, that is
a² = 1 ⇒ a = ± 1
a² = ⇒ a = ± 
Solutions are a = ± 1 , a = ±
Answer:$20250
Step-by-step explanation:
6.5% of 30000
6.5/100 x 30000
(6.5 x 30000)/100
195000/100=1950
First year depreciation 30000-1950=28050
Second year=28050-1950=26100
Third year=26100-1950=24150
Fourth year=24150-1950=22200
Fifth year=22200-1950=20250
Answer:
-6 -4 3 7
Step-by-step explanation:
This is right because if it goes- it goes lower and if it goes up on + it goes higher.
Answer:
The solutions are x = 1.24 and x = -3.24
Step-by-step explanation:
Hi there!
First, let´s write the equation:
log[(x² + 2x -3)⁴] = 0
Apply the logarithm property: log(xᵃ) = a log(x)
4 log[(x² + 2x -3)⁴] = 0
Divide by 4 both sides
log(x² + 2x -3) = 0
if log(x² + 2x -3) = 0, then x² + 2x -3 = 1 because only log 1 = 0
x² + 2x -3 = 1
Subtract 1 at both sides of the equation
x² + 2x -4 = 0
Using the quadratic formula let´s solve this quadratic equation:
a = 1
b = 2
c = -4
x = [-b± √(b² - 4ac)]/2a
x = [-2 + √(4 - 4(-4)·1)]/2 = 1.24
and
x = [-2 - √(4 - 4(-4)·1)]/2 = -3.24
The solutions are x = 1.24 and x = -3.24
Have a nice day!
When calculating accrued interest over several years that compounds annually, you must calculate a new principle each year, adding the accrued interest from the previous year. At the beginning of the new interest period, all the accrued interest is added to the principal which forms a new principle figure that the interest is then counted on.
