B) 4(x+2) + 8(x-1)
first we distribute the 4
4x+ 8 + 8(x-1)
now we distribute the 8
4x+8 + 8x-8
now we combine the 4x and the 8x because they have the same variable so they are like terms
12x +8-8
the 8 cancels out so
b) 12x
now c)
3(w-5) - 6(w+2)
distribute 3
3w-15 - 6(w+2)
now distribute -6
3w-15 -6w -12
combine 3w and -6w because they are like terms
-3w -15-12
now combine -15 and -12
answer to c: -3w-27
Answer:no
Step-by-step explanation: if they are all under the bus canopy they obviously take the bus make the data bias
Answer:
Given the information in the attached image;
The number of blue marbles is 5.
The total number of marbles is 10.
So, the probability of drawing blue as the first marble is;
![P_1=\frac{5}{10}=\frac{1}{2}](https://tex.z-dn.net/?f=P_1%3D%5Cfrac%7B5%7D%7B10%7D%3D%5Cfrac%7B1%7D%7B2%7D)
Also, since there is a replacement, the probability of drawing blue as the second marble is;
![P_2=\frac{5}{10}=\frac{1}{2}](https://tex.z-dn.net/?f=P_2%3D%5Cfrac%7B5%7D%7B10%7D%3D%5Cfrac%7B1%7D%7B2%7D)
the probability of drawing two blue marbles will be;
![\begin{gathered} P=P_1\times P_2 \\ P=\frac{1}{2}\times\frac{1}{2} \\ P=\frac{1}{4} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20P%3DP_1%5Ctimes%20P_2%20%5C%5C%20P%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%5Cfrac%7B1%7D%7B2%7D%20%5C%5C%20P%3D%5Cfrac%7B1%7D%7B4%7D%20%5Cend%7Bgathered%7D)
Therefore, the probability of drawing two blue marbles is;
![\frac{1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D)
Answer:
w x 3/4 = 35.7
Multiply both sides by 4
then, divide both sides by 3.
That is the answer. I assume you can take it from here.
A+b 5
------- = ---
2a-b 4
b 5
-------- = ---- => 9b = 5a+45 => b = (5a+45)/9
a+9 9
a+b 5
------- = --- => 4a+4b = 10a-5b
2a-b 4
9b = 6a
9(5a+45) =6a divide both sides by 3
3(5a+45) = 2a
15a +135 = 2a
13a = -135
a = -135/13 = -10,38
a= -10,38
9b=6a
3b=2a
3b = 2(-10,38)
3b = -20,76
b = -20,76/3
b = -6,92
hope this helped