F(x) = 3x + 1
let f(x) = y
y = 3x + 1
y - 1 = 3x
3x = y - 1
x = (y - 1)/3
From y = f(x), x = f⁻¹(y)
<span>x = (y - 1)/3
</span>
f⁻¹(y) = (y - 1)/3
f⁻¹(7) = <span>(7 - 1)/3 = 6/3 = 2
</span>f⁻¹(7)<span> = 2</span>
Expression 1 is 25
6*7-3^2*9+4^3
First simplify all the exponents to get 6*7 - 9 * 9 + 64 and then multiply the two numbers to get 42 - 81 + 64 and just add = 25
For expression 2, if you put the parantheses around (5+4)*2+6-2*2-1, it'll be 19
(5+4)*2+6-2*2-1
9*2+6-2*2-1
18+6-4-1
24-5
19
Remember order of operations
Answer:
Step-by-step explanation:
We can directly substitute y of the first equation to the second equation.
-2x - 1 = 3x - 16
5x = 15
x = 3
Substituting back to any of the two equations, we get y = -2(3)-1 = -7. If you check with the second equation, y = 3(3)-16 = -7 as well.
Therefore .
There is a hole at x=4 because the top factors to (x-4)(x+3)
in the end we get
f(x)=(x-4)(x+3)/(x-4)
the (x-4) cancels out and we get
f(x)=x+3
so baiscally f(x) is y=x+3 but with a hole at (4,7)
one way is to make a piecewise function defined as(x^2-x-12)/(x-4) for all x≠4 and defined as 7 for x=4
or, to just do f(x)=x+3
Answer:
12a + 3b.
Step-by-step explanation:
3(4a + b)
= 3 * 4a + 3 * b
= 12a + 3b.
Hope this helps!