Answer:
1 st questions answer is third option
while the answer of other question is 3
Step-by-step explanation:
2nd qestions explanation
put value of x=2 in equation as x-2=0 from equation
(2)^3-6(2)^2+a2+10=2
8-24+a2+10=0
2a=-18+24
2a=6
a=3
Answer: 24
Yes (x-2) is a factor. There is a remainder.
Step-by-step explanation:
2³-8•2²+14•2-4
8-8•4+14•2-4
0•4+14•2-4
0+14•2-4
14•2-4
28-4
24
Hope this helps if so plz mark me brainly
1/4(q-12) + 1/3(q+9), Then you must use the distributive property to get 1/q-3+1/3q+3. Then you simplify by combining like terms to get, 7/12q. You do not have any constants because they cross each other out.
If the list is of numbers of students in each music class, and if their median is supposed to be 19, then the blank must represent the number 18. The number must be lower than 20, and must be 19 when averaged with 20. Its value must be ...
2*19 - 20 = 18
_____
Altogether there are 6 numbers, so the median (19) will be the average of the middle two. Three of the numbers are more than 19, so the other three must be less than 19. The lowest of the upper 3 numbers is 20, so 19 must be the average of 20 and the missing number, which must be 18.
a cube, will be like the one in the picture below, with all equal sides, and therefore with a volume of x³.
![\bf \textit{volume of a cube}\\\\ V=x^3~~ \begin{cases} x=side's~length\\[-0.5em] \hrulefill\\ V=343 \end{cases}\implies 343=x^3\implies \sqrt[3]{343}=x\implies 7=x](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cube%7D%5C%5C%5C%5C%0AV%3Dx%5E3~~%0A%5Cbegin%7Bcases%7D%0Ax%3Dside%27s~length%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0AV%3D343%0A%5Cend%7Bcases%7D%5Cimplies%20343%3Dx%5E3%5Cimplies%20%5Csqrt%5B3%5D%7B343%7D%3Dx%5Cimplies%207%3Dx)
