2-[6÷2+{6×1/2+(7/2-3/2)}]
2 answers:
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The infinite geometric series is converges if |r| < 1.
We have r = 0.7 < 1, therefore our infinite geometric series is converges.
The sum S of an infinite geometric series with |r| < 1 is given by the formula :
We have:
Substitute:
Answer: c. Converges, 40.
SOLUTION OF TRIANGLE:
a i) use this formula, a² = b² + c² - 2bc cos A
a² = 8² + 10² - 2(8)(10) cos 60
= 84
a = 9.17 cm
UW = 9.17 cm
ii) use formula no 12 as attached :
sin
to solve this,
( sin 40 / 11 ) x 9.17 = 0.5358511
press shift sin on ur calc & the answer is :
FUNCTION
a) f(x) = x-3
g(x) = x²-1
i) f(6) = 6-3
substitute x with the value given in the bracket into equation
f(6) = 6-3
= 3
ii) to do inverse function use this formula :
u = f(x)
u = x - 3
and then make x as the subject
x = u + 3
after that, just change u to x
f-1(x) = x + 3
iii) f [ g(2) ] = f [ g(-2) ] = 0
g(2) = (2)² - 1 g(-2) = (-2)² - 1
= 4 - 1 = 4 - 1
= 3 = 3
f(3) = 3-3
= 0
a i) use this formula, a² = b² + c² - 2bc cos A
a² = 8² + 10² - 2(8)(10) cos 60
= 84
a = 9.17 cm
UW = 9.17 cm
ii) use formula no 12 as attached :
sin
to solve this,
( sin 40 / 11 ) x 9.17 = 0.5358511
press shift sin on ur calc & the answer is :
FUNCTION
a) f(x) = x-3
g(x) = x²-1
i) f(6) = 6-3
substitute x with the value given in the bracket into equation
f(6) = 6-3
= 3
ii) to do inverse function use this formula :
u = f(x)
u = x - 3
and then make x as the subject
x = u + 3
after that, just change u to x
f-1(x) = x + 3
iii) f [ g(2) ] = f [ g(-2) ] = 0
g(2) = (2)² - 1 g(-2) = (-2)² - 1
= 4 - 1 = 4 - 1
= 3 = 3
f(3) = 3-3
= 0