$H(x)=(x^2)+1\ \ \ \ and\ \ \ \ K(x)=-(x^2)+4. \\\\K(H)=-(H^2)+4=-(x^2+1)^2+4=4-(x^2+1)^2=\\\\.\ \ \ =2^2-(x^2+1)^2=(2-x^2-1)(2+x^2+1)=(1-x^2)(3+x^2)=\\\\.\ \ \ =(1-x)(1+x)(x^2-3\cdot i^2)=(1-x)(1+x)(x- \sqrt{3} \cdot i)(x+ \sqrt{3} \cdot i)\\\\ K(H)=0\ \ \ \ \Leftrightarrow\ \ \ (1-x)(1+x)(x- \sqrt{3} \cdot i)(x+ \sqrt{3} \cdot i)=0\\\\x=1\ \ \ \ or\ \ \ \ x=-1\ \ \ \ or\ \ \ \ x=\sqrt{3} \cdot i\ \ \ \ or\ \ \ \ x=-\sqrt{3} \cdot i\\\\Ans.\ e.$

4 0

3 years ago

Can I have somebody answer a few more of the questions that I need please and this one too?

tankabanditka [31]

Answer:

x > 22

Step-by-step explanation:

<em>Hey there!</em>

Well to solve,

52 - 3x < -14

we need to single out x

52 - 3x < -14

-52 to both sides

-3x < -66

Divide both sides by -3

x > 22

The < changes to > because the variable number is a - being divided.

<em>Hope this helps :)</em>

7 0

3 years ago

Read 2 more answers

Help is much needed pls. I can only put 15 points.

Alexxx [7]

Answer:

9.2

Step-by-step explanation:

first i added 5 + 3 = 8

then i did 1 1/5 + 8=9 1/5

6 0

3 years ago