<span>Assume that,
x^2+4x-5 = 0 .......(1)
Then,
x^2+4x-5 = 0
x^2+5x-1x-5 =0
x(x+5)-1(x+5) = 0
(x+5) (x-1) = 0
We get x=-5 and x=1
Sub x=-5 in equ (1)
(-5)^2+4(5)-5 = 0
-25+20-5 = 0
-25+25= 0
0 = 0
Sub x=1 in equ (1)
(1)^2+4(1)-5 = 0
1+4-5 = 0
5-5 = 0
0 = 0
Therefore x value is -5 and 1</span>
Answer:
8% or 0.08
Step-by-step explanation:
Probability of missing the first pass = 40% = 0.40
Probability of missing the second pass = 20% = 0.20
We have to find the probability that he misses both the passes. Since the two passes are independent of each other, the probability that he misses two passes will be:
Probability of missing 1st pass x Probability of missing 2nd pass
i.e.
Probability of missing two passes in a row = 0.40 x 0.20 = 0.08 = 8%
Thus, there is 8% probability that he misses two passes in a row.
Answer
-2.25 -13/5 1 3/5 -2.25
Step-by-step explanation:
Answer:
D = - 9
Step-by-step explanation:
D - 7 = -18
D - 7 + 7 = -18 + 7
D = -9
