3x + 4
2x + 5|6x² + 23x + 20
6x² + 15x
8x + 20
8x + 20
0
The answer is B.
(2n³ + 4n² - 7) + (-n³ - 8n - 2)
(2n³ - n³) + 4n² - 8n - 7 - 2
n³ + 4n² - 8n - 9
The answer is B.
2x² - 10x - 12
2(x²) - 2(5x) - 2(6)
2(x² - 5x - 6)
2(x² - 3x - 2x - 6)
2(x(x) - x(3) - 2(x) - 2(3))
2(x(x - 3) - 2(x - 3))
2(x - 2)(x - 3)
The answer is A.
4ab - 8a² + 12ac = 4ab - 8a² + 12ac = b - 2a + 3c
4a 4a 4a 4a
The answer is B.
Answer:
below
Step-by-step explanation:
that's the answer
let's recall that d = rt, distance = rate * time.
we know that Steve is twice as fast as Jill, so say if Jill has a speed or rate of "r", then Steve is traveling at 2r fast, now we know they both in opposite directions have covered a total of 120 miles, so if Jill covered "d" miles then Steve covered 120 -d, check the picture below.
![\begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Jill&d&r&2.5\\ Steve&120-d&2r&2.5 \end{array}~\hfill \begin{cases} d=2.5r\\[2em] 120-d=5r \end{cases} \\\\\\ \stackrel{\textit{substituting on the 2nd equation}}{120-2.5r=5r\implies 120=7.5r}\implies \cfrac{120}{7.5}=r\implies \stackrel{Jill's}{16=r}~\hfill \stackrel{Steve's}{32}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bmiles%7D%7Bdistance%7D%26%5Cstackrel%7Bmph%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20Jill%26d%26r%262.5%5C%5C%20Steve%26120-d%262r%262.5%20%5Cend%7Barray%7D~%5Chfill%20%5Cbegin%7Bcases%7D%20d%3D2.5r%5C%5C%5B2em%5D%20120-d%3D5r%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20on%20the%202nd%20equation%7D%7D%7B120-2.5r%3D5r%5Cimplies%20120%3D7.5r%7D%5Cimplies%20%5Ccfrac%7B120%7D%7B7.5%7D%3Dr%5Cimplies%20%5Cstackrel%7BJill%27s%7D%7B16%3Dr%7D~%5Chfill%20%5Cstackrel%7BSteve%27s%7D%7B32%7D)
Answer:
True
Step-by-step explanation:
If my eyes don’t fail me, it‘s true.
Sorry in advance if it’s wrong :<