$(x^2-3x+1)^{(2x^2+x-6)}=1\ we \ take \ the \ logarithm\\\\(2x^2+x-6)*ln(x^2-3x+1)=0\\\\(2x^2+x-6)=0\ or\ x^2-3x+1=1\\\\1)\\2x^2+x-6=0\\2x^2+4x-3x-6=0\\2x(x+2)-3(x+2)=0\\(x+2)(2x-3)\\x=-2\ or\ x=\dfrac{3}{2} \\\\2)\\x^2-3x=0\\x(x-3)=0\\x=0\ or\ x=3\\\\sol=\{-2,0,\dfrac{3}{2},3\}\\$

8 0

3 years ago

Can you please help with both? :)

QveST [7]

Answer: 1. 12 2. 120

Step-by-step explanation:

1. so basically 9/6=18/x

since they are similir the'd have the same ratio

sooo you just solve for x (multiply 6 by 18, and then divide by 9) and you get 12

2. same thing except you put 6/10=12/x, and solve it (multiply 10 by 12 and divide by 6) and u get 20

dats how i would do it but there are other ways too im sure :)

hope this helps

8 0

3 years ago

Compare the following ratios: (Use >, < or = sign in the place holders)

Zepler [3.9K]

Answer:

7. 2:3<__4:1

8. 1:2<__3:6

9. 3:5>__2:4

10. 6:7>__4:8

Step-by-step explanation:

4 0

3 years ago