The two factors of a quadratic equation can be multiplied to form the original equation. Let the missing term be a:
y² + 15y + 56 = (y + 7)(y + a)
y² + 15y + 56 = y² + (7 + a)y + 7a
We can now compare the coefficient of the like terms, either the one with y or the one without y, and find the value of a.
15 = 7 + a; a = 8
OR
56 = 7a; a = 8
Second factor is (y + 8)
Note: I am assuming you meant to say:
- Zoey has 99 beads instead of 99 years.
Answer:
Zoey needs 51 more beads to make 5 bangles.
Step-by-step explanation:
Total number of beads Zoey currently has = 99 beads
Beads need to make one bangel = 30 beads
so
Beads need to make 5 bangles = 30 × 5 = 150 beads
Since she has already 99 beads.
So, all we need s to subtract 99 beads from 150 beads.
Therefore,
The number of beads Zoey needs = 150 - 99
= 51 beads
Thus, Zoey needs 51 more beads to make 5 bangles.
Answer:
3:20
Step-by-step explanation:
4:25-15= 4:10
4:10-20=3:50
3:50-30=3:20
School ended at 3:20
1. a b c
2. a b c
3. a b c
4. a b c
5. a b c
then she eliminated 1 choice in 1 and 2, say as follows
1. b c
2. a b
3. a b c
4. a b c
5. a b c
Probability of answering correctly the first 2, and at least 2 or the remaining 3 is
P(answering 1,2 and exactly 2 of 3.4.or 5.)+P(answering 1,2 and also 3,4,5 )
P(answering 1,2 and exactly 2 of 3.4.or 5.)=
P(1,2,3,4 correct, 5 wrong)+P(1,2,3,5 correct, 4 wrong)+P(1,2,4,5 correct, 3 wrong)
also P(1,2,3,4 c, 5w)=P(1,2,3,5 c 4w)=P(1,2,4,5 c 3w )
so
P(answering 1,2 and exactly 2 of 3.4.or 5.)=3*P(1,2,3,4)=3*1/2*1/2*1/3*1/3*2/3=1/4*2/9=2/36=1/18
note: P(1 correct)=1/2
P(2 correct)=1/2
P(3 correct)=1/3
P(4 correct)=1/3
P(5 wrong) = 2/3
P(answering 1,2 and also 3,4,5 )=1/2*1/2*1/3*1/3*1/3=1/108
Ans: P= 1/18+1/108=(6+1)/108=7/108
Answer:
(a). 72.9%.
(b). 13.6 hr.
Step-by-step explanation:
So, we are given the following data or parameters or information which is going to assist us in solving this question/problem;
=> "A welder produces 7 welded assemblies during the first day on a new job, and the seventh assembly takes 45 minutes (unit time). "
=> The worker produces 10 welded assemblies on the second day, and the 10th assembly on the second day takes 30 minutes"
So, we will be making use of the Crawford learning curve model.
T(7) + 10 = T (17) = 30 min.
T(7) = T1(7)^b = 45.
T(17 ) = T1(17)^b = 30.
(T1) = 45/7^b = 30/17^b.
45/30 = 7^b/17^b = (7/17)^b.
1.5 = (0.41177)^b.
ln 1.5 = b ln 0.41177.
0.40547 = -0.8873 b.
b = - 0.45696.
=> 2^ -0.45696 = 0.7285.
= 72.9%.
(b). T1= 45/7^ - 045696 = 109.5 hr.
V(TT)(17) = 109.5 {(17.51^ - 0.45696 – 0.51^ - 0.45696) / (1 - 0.45696)} .
V(TT) (17) = 109.5 {(4.7317 - 0.6863) / 0.54304} .
= 815.7 min .
= 13.595 hr.
