SOS
Answer:
Completing the square is a method that will solve all quadratic equations. By completing the square on the general quadratic equation, we obtain a formula that is worth learning and that will solve all quadratic equations.
The quadratic formula does the same thing as completing the square. The formula reads:
The a, b and c terms are the coefficients of a polynomial where a is the coefficient of the squared term, b the linear term and c the constant when the equation is set equal to zero [ax squared plus bx plus c = 0].
Solving quadratics with this formula is then just as simple as substituting the numbers from an equation into these placeholder variables. Then, you need to simplify the expression to get the two solutions, one where you use the plus sign and the other the minus sign.
<em>Hope this helps!!</em>
Answer:
The constant of variation for the given quadratic equation is, 30
Step-by-step explanation:
One of the form of a quadratic equation is written as:
....[1]
where k is the coefficient and for this case the constant of variation.
In order to obtain the answer for the given equation, we write the given equation to the form above.
or
or
Comparing this equation with equation [1], to get the value of k;
k=30.
therefore, the constant of variation is, 30.
a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)
b. The inverse image of g(x) is
c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
<h3 /><h3>The domain of T</h3>Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0
⇒ x² ≥ 25
⇒ x ≥ ±5
⇒ -5 ≤ x ≤ 5.
<h3>Inverse image of f(x)</h3>
The inverse image of f(x) is f⁻¹(x) = ∛(x/3)
f : R → R defined by f(x) = 3x³
Let f(x) = y.
So, y = 3x³
Dividing through by 3, we have
y/3 = x³
Taking cube root of both sides, we have
x = ∛(y/3)
Replacing y with x we have
y = ∛(x/3)
Replacing y with f⁻¹(x), we have
So, the inverse image of f(x) is f⁻¹(x) = ∛(x/3)
<h3>Inverse image of g(x)</h3>
The inverse image of g(x) is
g : R+ → R defined by g(x) = ln(x).
Let g(x) = y
y = ln(x)
Taking exponents of both sides, we have
Replacing x with y, we have
Replacing y with g⁻¹(x), we have
So, the inverse image of g(x) is
The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
h : R → R defined by h(x) = x − 9
Let y = h(x)
y = x - 9
Adding 9 to both sides, we have
y + 9 = x
Replacing x with y, we have
x + 9 = y
Replacing y with h⁻¹(x), we have
So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
Learn more about inverse image of a function here: