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AlexFokin [52]
3 years ago
14

Multiply (x^2-5x+1)(4x-3x^2)

Mathematics
2 answers:
aleksley [76]3 years ago
6 0

x= 6 is the answer

Step-by-step explanation:

hope it helps

Andreas93 [3]3 years ago
6 0

Answer:

 -x • (x2 - 5x + 1) • (3x - 4)

Step-by-step explanation:

STEP

1: Equation at the end of step 1

 (((x2) -  5x) +  1) • (4x -  3x2)

STEP2:STEP

2:Pulling out like terms

 Pull out like factors :

4x - 3x2  =   -x • (3x - 4)

Trying to factor by splitting the middle term

 Factoring  x2 - 5x + 1

The first term is,  x2  its coefficient is  1 .

The middle term is,  -5x  its coefficient is  -5 .

The last term, "the constant", is  +1

Step-1 : Multiply the coefficient of the first term by the constant   1 • 1 = 1

Step-2 : Find two factors of  1  whose sum equals the coefficient of the middle term, which is   -5 .

     -1    +    -1    =    -2

     1    +    1    =    2

Observation : No two such factors can be found !!

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Given: △ABC, AB = CB AK ⊥ BC , MC ⊥ AB Prove: AK = MC
MatroZZZ [7]

By using congruence of two triangles AKC and CMA we proved that

AK = MC

Step-by-step explanation:

Lets revise the cases of congruence  

  • SSS  ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
  • SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and   including angle in the 2nd Δ
  • ASA ⇒ 2 angles and the side whose joining them in the 1st Δ  ≅ 2 angles and the side whose joining them in the 2nd Δ
  • AAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles   and one side in the 2nd Δ
  • HL ⇒ hypotenuse leg of the 1st right Δ ≅ hypotenuse leg of the 2nd right Δ

In Δ ABC

∵ AB = BC

∴ Δ ABC is an isosceles triangle

∴ ∠BAC ≅ m∠BCA ⇒ base angles of isoscelesΔ

∵ AK ⊥ BC

∴ m∠AKC = 90°

∵ MC ⊥ AB

∴ m∠CMA = 90°

∴ ∠AKC ≅ ∠CMA

In the two Δs AKC and CMA

∵ ∠BAC ≅ m∠BCA

∵ ∠AKC ≅ ∠CMA

∵ AC ≅ CA

∴ Δ AKC and Δ CMA are congruent by AAS postulate

∴ AK ≅ CM

By using congruence of two triangles AKC and CMA we proved that

AK = MC

Learn more:

You can learn more about congruence in brainly.com/question/3202836

#LearnwithBrainly

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