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Sergeu [11.5K]
4 years ago
11

Lines I and m are parallel. Find m23 if m25 = 38 and m26=62. 100 62 80 O 38

Mathematics
2 answers:
kotegsom [21]4 years ago
6 0
The answer to yourrrrr question is O38
snow_tiger [21]4 years ago
5 0

Answer:

∠3=80°

Step-by-step explanation:

∠5=∠2 (alternate angle)

∠5=38

so ∠2=38

∠6=62

∠3+∠6+∠2=180

∠3+62+38=180

∠3=180-100=80

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Part-A:-

<h3>(f×g)=</h3>

\\ \sf\longmapsto x^2+3x-70(4x-12)

\\ \sf\longmapsto 4x^2+12x-280-12x^2+36x+840

\\ \sf\longmapsto -8x^2+48x+560

Now

\\ \sf\longmapsto (f.g)(3)

\\ \sf\longmapsto -8(3)^2+48(3)+560

\\ \sf\longmapsto -72+144+560

\\ \sf\longmapsto 72+560

\\ \sf\longmapsto 632

Part-B:-

\\ \sf\longmapsto f(3)

\\ \sf\longmapsto (3)^2+3(3)-70

\\ \sf\longmapsto 9+9-70

\\ \sf\longmapsto 18-70

\\ \sf\longmapsto -52

Now

\\ \sf\longmapsto g(f(3))

\\ \sf\longmapsto g(-52)

\\ \sf\longmapsto 4(-52-3)

\\ \sf\longmapsto 4(-55)

\\ \sf\longmapsto -220

Part:-C

\\ \sf\longmapsto g(-2)

\\ \sf\longmapsto 4(-2-3)

\\ \sf\longmapsto 4(-5)

\\ \sf\longmapsto -20

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\\ \sf\longmapsto f(g(-2))

\\ \sf\longmapsto f(-20)

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\\ \sf\longmapsto 400-130

\\ \sf\longmapsto 270

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  • g×f=f×g

\\ \sf\longmapsto (g.f)(-2)

\\ \sf\longmapsto -8(-2)^2+48(-2)+560

\\ \sf\longmapsto 32-96+560

\\ \sf\longmapsto -64+560

\\ \sf\longmapsto 496

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