Plz help me ill give you brainlist

Mathematics

1 answer:

Ivan3 years ago

7 0

So, plug in what you're given and solve
x is -2
(3x) * 9
(3 (-2)) * 9
-6 * 9
-54

then
27(-2)
-54
For the first it's the same

for the second
x is 0
(3x) * 9
(3 (0)) * 9
0 * 9
0

27x
27(0)
0
For the second it's the same

x is 4
(3x) * 9
(3(4)) * 9
12 * 9
108

27x
27(4)
108

The given expressions are equivalent. This was proved in the diagram.

Hope this helps :)

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Example of an inverse operation

Alexxx [7]

Addition is the inverse of subtraction and multiplication is the inverse of division.

Ex: 1+1 = 2, 2-2 = 0

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3 years ago

Solve equation.<br> k+3=14

lorasvet [3.4K]

To solve the equation, subtract 3 from both sides. The solution is k = 11.

A linear equation is an equation of which the highest degree of its variables is 1.

Steps to solve linear equations:

Simplify the expressions. If there are parentheses, remove them by multiplying the corresponding terms.
Combine the same terms.
Isolate the variable on one side by using subtraction of addition.
Use division or multiplication to find the value of the variable.

Recall that if we subtract or add the same number to both sides of an equation, it does not change the equality.

The given problem:

Solve k + 3 = 14.

The variable is k. Subtract 3 from both sides:

k + 3 - 3 = 14 - 3

k = 11 (solved)

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1 year ago

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

Anon25 [30]

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 $(-\infty,5]$ .