Answer:
First, find tan A and tan B.
cosA=35 --> sin2A=1−925=1625 --> cosA=±45
cosA=45 because A is in Quadrant I
tanA=sinAcosA=(45)(53)=43.
sinB=513 --> cos2B=1−25169=144169 --> sinB=±1213.
sinB=1213 because B is in Quadrant I
tanB=sinBcosB=(513)(1312)=512
Apply the trig identity:
tan(A−B)=tanA−tanB1−tanA.tanB
tanA−tanB=43−512=1112
(1−tanA.tanB)=1−2036=1636=49
tan(A−B)=(1112)(94)=3316
kamina op bolte
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1834 bfc I believe that’s the right answer
Answer:
g(-1 )=-1 and g(2)+g(1)=7
Step-by-step explanation:
If g(x) = x^3+x^2-x-2 find g(-1)
if we find g(-1)
we substitute all the x's in the function with -1
-1^3+-1^2-(-1)-2
-1^3 = -1
-1^2 = 1
-1+1+1-2
(two minuses make a plus)
-1+1 = 0
0+1 = 1
1-2 = -1
if x=-1, g(-1) is -1
g(2)+g(1)
substitute the x's in the function with 2 and 1 and add your results
2^3+2^2-2-2
2^3 = 8, 2^2 = 4
8+4-2-2
8+4= 12, 12-2 = 10, 10-2 = 8
g(2)=8
g(1) now
1^3 + 1^2-1-2
1^3=1, 1^2 = 1
1+1-1-2
1+1 = 2, 2-1 = 1, 1-2 = -1
g(3) = -1
g(2) (which equals 8) + g(3) (which equals -1) =
8+(-1) = 7
g(2)+g(3)=7
Example question: There are 38 students in a baseball club. The coach needs to split them up into equal groups of 4. How many groups can be made? How many kids are left over?
Answer: You would make groups of 4 to get your quotient (9). Then whatever is NOT in a group is your left over.
Answer:
SO THE 1309 IS THE NEW 1 1(239) THE ANSWER OF THAT IS IN THE LINK MP MODULE .COM
