Need help finding the limits

Korvikt [17]

Answer:

x= -1

Step-by-step explanation:

1. First, you need to combine like terms. Constants with constants and variables with variables. Since there is only one variable, there is nothing to combine it with so nothing happens to it. The constants, on the other hand, can combine.

2. You have to subtract 28 from 28 and 14 so that +28 can cancel out on the left side and combine on the right.

3. That will leave you with 14x= -14.

4. Now you need to get x on its own so you need to divide 14 (opposite of multiplication to cancel out) on both sides.

5. That will leave you with x= -1. That is your final answer.

Hope this helps! Please vote me as Brainliest. Let me know if you need more of an explanation. Remember, cheating is never the answer. Try your best before you seek answers. Hope it's not too late. Good luck with whatever you are doing! Have an amazing day!

8 0

3 years ago

A box holds 8 red and 4 blue beads. Three beads are taken from the box and not replaced. Determine: The probability that all thr

zubka84 [21]

Answer: $\dfrac{14}{55}, \dfrac{13}{55}$

Step-by-step explanation:

Given

There are 8 red and 4 blue beads

Three beads were taken from the box

(i)No of ways of choosing three red beads out of 12 beads is $^8C_3$

Total no of ways $^{12}C_3$

Probability is

$P=\dfrac{^8C_3}{^{12}C_3}=\dfrac{8\times 7\times 6}{12\times 11\times 10}=\dfrac{2\times 7}{11\times 5}\\P=\dfrac{14}{55}$

(ii)atleast two beads refers to minimum two beads

There can be two possibilities (2B,1 R), (3B, 0R)

$\Rightarrow P=\dfrac{^4C_2\times ^8C_1}{^{12}C_3}+\dfrac{^4C_3}{^{12}C_3}\\\\\Rightarrow P=\dfrac{6\times 8}{220}+\dfrac{4}{220}=\dfrac{52}{220}\\\\\Rightarrow P=\dfrac{13}{55}$