Answer: x = ¹/₂ ± √⁸¹
------------
2
Step-by-step explanation:
First write out the equation
x² - x - 20
Now we now write the equation by equating to 0
x² - x - 20 = 0
We now move 20 to the other side of the equation. So
x² - x = 20,
We now add to both side of the equation square of the half the coefficient of the (x) and not (x²) which is (1) . So, the equation now becomes
x² - x + ( ¹/₂ )² = 20 + ( ¹/₂ )²
x² - ( ¹/₂ )² = 20 + ¹/₄
( x - ¹/₂ )² = 20 + ¹/₄, we now resolve the right hand side expression into fraction
( x - ¹/₂ )² = ⁸¹/₄ when the LCM is made 4
Taking the square root of both side to remove the square,we now have
x - ¹/₂ = √⁸¹/₄
x - ¹/₂ = √⁸¹/₂
Therefore,
x = ¹/₂ ± √⁸¹
-----------
2
Answer:
-5 is the value
Step-by-step explanation:
47-42=-L
5=-L
L=-5
Answer:
- 8p - 80
Step-by-step explanation:
Given
4(- 15 - 3p) - 4(- p + 5) ← distribute both parenthesis
= - 60 - 12p + 4p - 20 ← collect like terms
= - 8p - 80
Answer:
Difference between upper and lower limits is : 1,816
Step-by-step explanation:
A CI (confidence interval ) for t student distribution is:
( μ₀ - t(α/2)* s/√n ; μ₀ + t(α/2)* s/√n )
Where:
μ₀ is the mean and s the standard deviation of the dstribution
n size of the sample
CI = 90 % means α = 10 % α = 0,1 α/2 = 0,05
and degree of freedom df = n - 1 df = 40
From t student table we get:
tα/2 = 1,6839
Then:
t(α/2)* s/√n = 1,6839* 3,41/√40
t(α/2)* s/√n = 0,908
8,73 - 0,908 = 7,822
8,73 + 0,908 = 9,638
CI (90%) = ( 7,822 ; 9,638 )
Difference between upper and lower cut-offs points is:
Δ = 1,816
Answer:
(- 5, 1 )
Step-by-step explanation:
- 6x - 14y = 16 → (1)
- 2x + 7y = 17 → (2)
Multiplying (2) by - 3 and adding to (1) will eliminate the x- term
6x - 21y = - 51 → (3)
Add (1) and (3) term by term to eliminate x
0 - 35y = - 35
- 35y = - 35 ( divide both sides by - 35 )
y = 1
Substitute y = 1 into either of the 2 equations and solve for x
Substituting into (1)
- 6x - 14(1) = 16
- 6x - 14 = 16 ( add 14 to both sides )
- 6x = 30 ( divide both sides by - 6 )
x = - 5
solution is (- 5, 1 )
