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Klio2033 [76]
3 years ago
7

Please help (URGENT) (im slow)

Mathematics
1 answer:
MAVERICK [17]3 years ago
6 0

Answer:

x=8

Step-by-step explanation:

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y = 2x + 13

Step-by-step explanation:

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Please help and show work!!
Jlenok [28]

A(1, 1), B(7, 1), C(1, 9)

AB = 7 - 1 = 6

AC = 9 - 1 = 8

Use the Pythagorean theorem:

AB² + AC² = BC²

Substitute:

BC² = 6² + 8²

BC² = 36 + 64

BC² = 100 → BC = √100 → BC = 10

The perimeter of ΔABC:

P = 6 + 8 + 10 = 24

<h3>Answer: 24 units</h3>
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I dont understand this!
DaniilM [7]

Answer:perimeter  

Step-by-step explanation:20+20+4=44 and u multiply 3/5x times 2 it equals 6/10x+44

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The shape of this graph is best described as 2012 housing prices
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2 years ago
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
siniylev [52]

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,

  • angle A is the common angle, so \angle MAP\cong \angle DAC by reflexive property;
  • 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}

4 0
3 years ago
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