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lakkis [162]
2 years ago
10

3 (h−9) + 2h = h − 27 + 4h

Mathematics
2 answers:
Sergeu [11.5K]2 years ago
6 0

\\ \sf\longmapsto 3(h-9)+2h=h-27+4h

\\ \sf\longmapsto 3h-27+2h=5h-27

\\ \sf\longmapsto 5h-27=5h-27

\\ \sf\longmapsto 5h-5h=-27+27

\\ \sf\longmapsto 0=0

The expression has no solutions

kenny6666 [7]2 years ago
5 0

Answer:

  • h - any real number

Step-by-step explanation:

  • 3 (h−9) + 2h = h − 27 + 4h
  • 3h - 27 + 2h = 5h - 27
  • 5h - 27 = 5h - 27
  • 0 = 0

h = any number

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In the trapezoid ABCD (AB∥CD) point M∈AD, so that AM:MD=3:5. Line L ∥AB and going through point M intersects diagonal AC and leg
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Answer:

\dfrac{AP}{PC}=\dfrac{3}{5}

\dfrac{BN}{CN}=\dfrac{3}{5}

Step-by-step explanation:

Consider triangles AMP and ADC. In these triangles,

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  • angles AMP and ADC are congruent as corresponding angles when two parallel lines MP and CD are cut by transversal AD.

Hence, triangles AMP and ADC are similar by AA similarity theorem.

Similar triangles have proportional corresponding sides, thus

\dfrac{AM}{AD}=\dfrac{AP}{AC}\\ \\\dfrac{3x}{3x+5x}=\dfrac{AP}{AC}\\ \\\dfrac{AP}{AC}=\dfrac{3}{8}\Rightarrow AP=\dfrac{3}{8}AC\\ \\PC=AC-AP=AC-\dfrac{3}{8}AC=\dfrac{5}{8}AC,

so

\dfrac{AP}{PC}=\dfrac{\frac{3}{8}AC}{\frac{5}{8}AC}=\dfrac{3}{5}

Consider triangles ACB and PCN. In these triangles,

  • angle C is the common angle, so \angle ACB\cong \angle PCN by reflexive property;
  • angles ABC and PCN are congruent as corresponding angles when two parallel lines PN and AB are cut by transversal BC.

Hence, triangles ACB and PCN are similar by AA similarity theorem.

Similar triangles have proportional corresponding sides, thus

\dfrac{CP}{AP}=\dfrac{CN}{CB}\\ \\\dfrac{5x}{3x+5x}=\dfrac{CN}{CB}\\ \\\dfrac{CN}{CB}=\dfrac{5}{8}\Rightarrow CN=\dfrac{5}{8}CB\\ \\BN=BC-CN=BC-\dfrac{5}{8}BC=\dfrac{3}{8}BC,

so

\dfrac{BN}{CN}=\dfrac{\frac{3}{8}BC}{\frac{5}{8}BC}=\dfrac{3}{5}

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