Solve: 2a-3=9 Solve: -3(b+2)= -9

2 answers:

8 0

2a - 3 = 9......
the first thing you would have to do is add a 3 on both sides of the equation.
2a - 3 =9        ------>    2a = 12
  +3      +3
Next, divide the 2 out from the a.
2a = 12    ---------> a = 6
/2       2/
_________________________________________________
-3(b + 2) = -9
multiply the -3 in the parentheses

-3(b + 2) = -9 -------> -3b - 6 = -9
Add a 6 to both sides
-3b - 6 = -9 --------------> -3b = -3
     +6      +6
Lastly divide -3 from both sides
-3b = -3 ----------------------> b = 1
/-3     /-3

Softa [21]2 years ago

3 0

The Solution are : a=3.
b=1.

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The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

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We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

If we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

Let us know find its inverse.

$y=(x-5)^{2}+2$

Upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

Let us now solve this function for y.

$(y-5)^{2}=x-2\\
y-5=\pm \sqrt{x-2}\\
y=5\pm \sqrt{x-2}\\$

Hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .