If sinα=cosß it means that ß=90-α. This also means that α+ß=90 so we can say:
4k-22+6k-13=90
10k-35=90
10k=125
k=12.5°
...
The above identity comes up a lot, and is one worth committing to memory for sure.
sinα=cos(90-α) or vice versa cosα=sin(90-α)
Answer:
17
Step-by-step explanation:
Step-by-step explanation:
6+[11-{10+2-(8×2+1-3×4)}]
6+[11-{10+2-(8×2+1-3×4)}]6+[11-{10+2-(12)}
6+[11-{10+2-(8×2+1-3×4)}]6+[11-{10+2-(12)} ]6+[11-[10+2-12}]
6+[11-{10+2-(8×2+1-3×4)}]6+[11-{10+2-(12)} ]6+[11-{10+2-12}]6+[11-{0}]
6+[11-0]
6+[11]
6+11
17 Ans.
Answer
6ac−2b−2
Step-by-step explanation:
Let's simplify step-by-step.
4ac+3−2b+2ac−5
=4ac+3+−2b+2ac+−5
Combine Like Terms:
=4ac+3+−2b+2ac+−5
=(4ac+2ac)+(−2b)+(3+−5)
=6ac+−2b+−2
Answer: 6/12 are white, 3/12 are colored and 3/12 are albino.
Step-by-step explanation: If the horses are white and their parents are ccww (albino) and CCWw (white horse), according to Mendel's premises, they both must be CcWw, since the crossing provides one C from one parent and other c from the other parent, one W and the other w. Using Mendel's chess and the principle of independent segregation, the crossing between CcWw results in the following fenotypical ratio:
1/16 CCWW (lethal)
2/16 CCWw (white)
2/16 CcWW (lethal)
4/16 CcWw (white)
1/16 CCww (normal)
2/16 Ccww (normal)
2/16 ccWw (albino)
1/16 ccWW (lethal)
1/16 ccww (albino)
Excluding the 4 individuals that have the lethal locus, we have 6/12 that are white (2/12 + 4/12) and 3/12 (1/12 + 2/12) that are colored. Also, there are 3/12 of albino individuals as well.
Answer:
1- (-1,4)
2- (3,8)
Step-by-step explanation:
PRETTY SURE
