What is nine one fourth times eight and four fiths
2 answers:
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Answer:
see below
Step-by-step explanation:
some of your answers that you currently have are wrong, I'll note those mistakes below
- when factoring (ie 5x+15) only factor out things that can divide both numbers into a whole number ratio
5x+15 = 5(x+3), not (x+3)(x+5)
ie
we see that 10x can divide the numerator in a whole number ratio
= 10x(x+2), not (x+2)(x+10)
second mistake: the first binomial expansion is incorrect.
you have the expansion formula right, but you added terms wrong, go look at it again
3. x^2+3x+2/ x^2+5x+6
(x+1)(x+2)/(x+3)(x+2)
(x+1)/(x+3)
4. (x^2+6x+8)/(x^2-16)
(x+4)(x+2)/(x+4)(x-4)
(x+2)/(x-4)
5. we can't simplify that any more, x and y are different variables so therefore we cannot cross out stuff on numerator and denominator
6. (x^4y^6)^2
(x^4y^6)(x^4y^6) =
remember that (x^a)(x^b) = x^(a+b)
or remember that (x^a)^b = x^ab
Answer:
39 is the 13th term
Step-by-step explanation:
Okay so essentially the is just a substitute for x.
See how in the equation they put n into the x spot:
They changed the variable names:
Now we know what the n actually means, plus when they ask about the terms youre trying to find it's based off of that n:
When they say, 13 term, they mean which would be the 13th answer in the series. So the first term would be based off of
So whatever number is in the n position by a is what you'll be plugging into the x spot.
These are all the answers leading up to the 13th term they asked for, and you can continue thus same equation with whatever number you can put into the n space.