Answer:
D I'm sure....................... :D
Answer:
3(2x^2-3x+14)
Step-by-step explanation:
6x^2-9x+42
All three terms have a common factor of 3
3(2x^2-3x+14)
Now let's focus on 2x^2-3x+14 and bring down the factor 3 later
so a=2
b=-3
c=14
Let's try to find two factors for ac that multiply to be a*c and add up to be b.
ac=28
b=-3
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ac=7(4)=14(2)=8(2)
Even if I made these pairs with both negatives nothing would give me -3
So you can only go as far as 3(2x^2-3x+14)
Here is another thing to help you if you have ax^2+bx+c and b^2-4ac<0 then it can't be factored (over reals)
Answer:
Step-by-step explanation:
Or, if you mean (r+3Q)/h=t:
1,045 to the nearest thousands would be <u>1,000</u> because you must round down since there are no numbers close enough to round up.
Could you send a picture of the question