$2n^2+3n-9= \\\\ a=2 \\ b=3 \\ c=-9 \\\\ \Delta= 3^2-4*2*(-9)=9+72\to\boxed{81} \\\\ x_1;x_2=\frac{-3 +\-\sqrt{81}}{2*2}= \frac{-3+/- 9}{4} \\\\ x_1=\frac{-3+9}{4}=\frac{6}{4}\to\boxed{\frac{3}{2}} \\\\ x_2=\frac{-3-9}{4}=\frac{-12}{4}\to\boxed{-4} \\\\ 2n^2+3n-9=(x-\frac{3}{2})(\x+4)$

7 0

3 years ago

What is the inverse of f(x) = sqft x+ 3

grin007 [14]

Answer:

$f^{-1} (x)=x^{2} -3$

Step-by-step explanation:

Given the function

$f\left(x\right)=\sqrt{x+3}$

We can find the inverse function by exchanging x and y, and then solve for y.

$y=\sqrt{x+3}$

Replace x with y.

$x=\sqrt{y+3}$

square both sides

$x^2=y+3$

$y=x^2-3$

Hence,

$f^{-1} (x)=x^{2} -3$

8 0

3 years ago

Somebody, please help

SashulF [63]

F(x) can be written as:
f(x) = Asin(2x); where A is the amplitude and the period of the function is half that of a normal sin function.
f(π/4) = 4
4 = Asin(2(π/4))
4 = Asin(π/2)
A = 4
Amplitude of g(x) = 1/2 * amplitude of f(x)
A for g(x) = 2
g(x) = 2sin(x)

5 0

3 years ago