Find the missing numerator 5x+1 - 31−x - 7x−1x2−1 = ?x+1.
-1
x - 1
3(1−5x)x+1
1
3(1 - 5x)
Correct answer is D
5x+1−31−x−7x−1x2−1=?x+1
5(x−1)−3(−(x+1))−7x−1x2−1
= 5x−5+3x+3−7x−1x2−1
= x−1x2−1
= x−1(x−1)(x+1)
= 1x+1.
The numerator = 1.
500
2 log10 5
10
25
log105 x 10100
Correct answer is E
102 + log105 = log10 10100 + log105
= log105 x 10100
1.03
2.31
3.69
10.5
25
Correct answer is B
log 10.5 = log 212
= log 21 - log 2
= log(3 x 7) - log 2
= log 3 + log 7 - log 2
= 1.10 + 1.90 - 0.69
= 3 - 0.69
= 2.31
Solve the equation for the positive values of θ less than 360o. 3 tan θ + 2 = -1
135o or 315o
45o or 135o
315o or 180o
315v + 45o
360o or 315o
Correct answer is A
3 tan θ + 2 = -1
3 tan θ −33 = -1
θ = tan -1(-1)
θ = 360o - 45o
= 315o
θ = 180 - 45o = 135o
188.57cm2
1320cm2
188cm2
188.08cm2
10cm2
Correct answer is A
S = curved surface area = πrL
= 227 x 6 x 10
= 188.57cm2