An equation does not always have a solution. Sometimes, an equation can have a variable that does not equal the remaining value on the other side. For example, root negative x could never equal a number squared.
√-x ≠ 14²
This equation cannot have a solution because you can't square any number to get a negative solution. Therefore, x= no solution.
Answer:
Hope this helped.
A brainliest is always appreciated.
The only equation that works with x=-6 is A
The solution to the composite function f(g(x)) is 9x² - 78x + 165.
<h3>
What is composite function?</h3>
A composite function is generally a function that is written inside another function.
Function composition is an operation that takes two functions f and g, and produces a function h = g ∘ f such that h(x) = g.
From the given composite function, the solution is determined as follows;
to solve for f(g(x)), we use the following methods.
f(x) = x² + 2x - 3, g(x) = 3x - 14
f(g(x)) = (3x - 14)² + 2(3x - 14) - 3
= 9x² - 84x + 196 + 6x - 28 - 3
= 9x² - 78x + 165
Thus, the solution to the composite function f(g(x)) is 9x² - 78x + 165.
Learn more about composite function here: brainly.com/question/10687170
#SPJ1
The complete question is below:
F(x) =x2+2x-3 g(x)=3x-14, find f(g(x))
Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.
