(f + g)(x) means f(x) + g(x)
So, you just input the function of f(x) and g(x) altogether, then simplify it.
(f + g)(x)
= f(x) + g(x)
= 2x² + 1 + x² - 7
= 2x² + x² + 1 - 7
= 3x² - 6
The answer is b
Step-by-step explanation:
I guess method 1 means to deal with whole factors.
x + 5 = (x - 2)(x + 5)
for (x + 5) <> 0 we can divide both sides by this factor :
1 = x - 2
x = 3
for the second solution we deal with
x + 5 = 0
x = -5
so, for x = -5 and x = 3 both functions deliver the same output, and these are the intersection points.
method 2 : we multiply the expression out and solve it then
x + 5 = (x - 2)(x + 5)
x + 5 = x² + 5x - 2x - 10 = x² + 3x - 10
0 = x² + 2x - 15
the general solution to such a square equation is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = 2
c = -15
x = (-2 ± sqrt(2² - 4×1×-15))/(2×1) =
= (-2 ± sqrt(4 + 60))/2 = (-2 ± sqrt(64))/2 = (-2 ± 8)/2 =
= -1 ± 4
x1 = -1 + 4 = 3
x2 = -1 - 4 = -5
and you get the 2 solutions again. as expected, they are the same as with method 1, of course.
Answer:
1-i and -1+i
Step-by-step explanation:
We are to find the square roots of 
. First, convert from Cartesian to polar form:
Next, use the formula ![\displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%5Bn%5D%7Br%7D%5Cbiggr%5B%5Ccis%5Cbiggr%28%5Cfrac%7B%5Ctheta%2B2%5Cpi%20k%7D%7Bn%7D%5Cbiggr%29%5Cbiggr%5D)
where 
to find the square roots:
<u>When k=1</u>
<u />![\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%5B2%5D%7B2%7D%5Cbiggr%5Bcis%5Cbiggr%28%5Cfrac%7B%5Cfrac%7B3%5Cpi%7D%7B2%7D%2B2%5Cpi%281%29%7D%7B2%7D%5Cbiggr%29%5Cbiggr%5D)
![\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%7B2%7D%5Cbiggr%5Bcis%5Cbiggr%28%5Cfrac%7B3%5Cpi%7D%7B4%7D%2B%5Cpi%5Cbiggr%29%5Cbiggr%5D)
<u>When k=0</u>
<u />![\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%5B2%5D%7B2%7D%5Cbiggr%5Bcis%5Cbiggr%28%5Cfrac%7B%5Cfrac%7B3%5Cpi%7D%7B2%7D%2B2%5Cpi%280%29%7D%7B2%7D%5Cbiggr%29%5Cbiggr%5D)
Thus, the square roots of -2i are 1-i and -1+i
Answer:C.11 in
explanation:Hope I helped:)
