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Nataliya [291]
4 years ago
13

QF Q1.) Solve the triangle.

Mathematics
2 answers:
Norma-Jean [14]4 years ago
4 0
C=180-A-B=180-69-40=71deg.

by sine rule

b/sinB=a/sinA
b=a/sinA*sinB
=15/sin(69)*sin(40)
= 10.33

c/sinC=a/sinA
c=a/sinA*sinC
=15/sin(69)*sin(71)
= 15.19

sergiy2304 [10]4 years ago
3 0

Sum of interior angles in a triangle is 180 deg.


so angle C = 180 - A - B


=180 - 69 -  40


=71deg.



Use the sine rule in a triangle - sinA / a = sinB / b


b = a * sinB / sinA


=15 * sin(40) / sin(69)


=10.3



Similarly sinC / c = sinA / a


c = a * sinC / sinA


=13 * sin(71) / sin(69)


=15.2


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f(x)=x3−5

Replace f(x)

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y=x3−5

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y3=5+x

Take the cube root of both sides of the equation to eliminate the exponent on the left side.

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Replace the y

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f−1(x)=3√5+x

Set up the composite result function.

f(g(x))

Evaluate f(g(x))

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Remove parentheses around 3√5+x

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f(3√5+x)=3√5+x3−5

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f(3√5+x)=5+x−5

Simplify by subtracting numbers.

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Subtract 5

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f(3√5+x)=x+0

Add x

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f(3√5+x)=x

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f−1(x)=3√5+x

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