Answer:
Now, 3x 2+x+5⩾0
This is because b² −4ac=1−4×5×3
=−59 (roots are imaginary)
3x²+x+5=(x−3)² =x²+9−6x and x−3⩾0
2x²+7x−4=0
2x²+8x−x−4=0
2x(x+4)−1(x+4)=0
(2x−1)(x+4)=0
x= 1/2 and−4 but x≥3
∴ No solution.
lol hehehe
Answer:
1200
Step-by-step explanation:
4x6= 24
24x 5= 120
120x 10= 1200
1+2+3= 6 (even)
2+3+4= 9 (odd)
it can be both
123 - Use PEMDAS but since you have both division and multiplication you can just go from left to right
The domain is all the x-values you can use with this function. When dealing with a function like this, you need to ask yourself, what would make the bottom equal to 0? Whatever those numbers are, you do NOT want them in your domain.
A fundamental rule is that you cannot divide by 0 or have 0 as a denominator.
So, you solve x^2+6x=0 to find that x=0 or x=-6 would make the denominator 0.
The domain is everything BUT those two values, because they essentially break your function.
