Answer:
Therefore the two values required are 2.08 and -11.08.
Step-by-step explanation:
i) f(x) =
+ 10x - 12 is a quadratic equation and a quadratic equation will have two roots whose values when substituted in to the given quadratic equation will give the value.
ii) The question is therefore essentially asking us to find the roots of the given quadratic equation.
This can be done by equating the given quadratic equation to zero and then solving the equation for its roots.
∴
+ 10x -12 = 0 ⇒ x = (-10 ±
) ÷ 2
= (-10 ± 12.166)÷2 = 2.08, -11.08
Therefore the two values required are 2.08 and -11.08
Answer:
x=-14y
Step-by-step explanation:
9x+y=-5
x+y=-5-9
x+y=-14
x=-14y
Answer:
x = 5/39
, y = 539/39
Step-by-step explanation:
Solve the following system:
{y - 2.5 x = 13.5
12.25 x - y = -12.25
In the first equation, look to solve for y:
{y - 2.5 x = 13.5
12.25 x - y = -12.25
y - 2.5 x = y - (5 x)/2 and 13.5 = 27/2:
y - (5 x)/2 = 27/2
Add (5 x)/2 to both sides:
{y = 1/2 (5 x + 27)
12.25 x - y = -12.25
Substitute y = 1/2 (5 x + 27) into the second equation:
{y = 1/2 (5 x + 27)
1/2 (-5 x - 27) + 12.25 x = -12.25
(-5 x - 27)/2 + 12.25 x = 12.25 x + (-(5 x)/2 - 27/2) = 9.75 x - 27/2:
{y = 1/2 (5 x + 27)
9.75 x - 27/2 = -12.25
In the second equation, look to solve for x:
{y = 1/2 (5 x + 27)
9.75 x - 27/2 = -12.25
9.75 x - 27/2 = (39 x)/4 - 27/2 and -12.25 = -49/4:
(39 x)/4 - 27/2 = -49/4
Add 27/2 to both sides:
{y = 1/2 (5 x + 27)
(39 x)/4 = 5/4
Multiply both sides by 4/39:
{y = 1/2 (5 x + 27)
x = 5/39
Substitute x = 5/39 into the first equation:
{y = 539/39
x = 5/39
Collect results in alphabetical order:
Answer: {x = 5/39
, y = 539/39
F(x) = ln (6-x); find the domain.
The domain of the "ln" function is (0, infinity), or "x is greater than zero."
Thus, the domain of the given function is "(6-x) is greater than zero."
Solve this: 6-x > 0. Add x to both sides. Then 6 > x, or x < 6 (answer)
Answer:
13
Step-by-step explanation: