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IRINA_888 [86]
3 years ago
5

The perimeter of a triangle is 29. One side of a triangle is 5 less than twice the side of the shortest length. The third side i

s 6 more than the shortest side. What are the lengths of the triangle?
Mathematics
2 answers:
jekas [21]3 years ago
6 0

ASSUMPTION:

Let the shortest length be x.

ANSWER:

Perimeter of Δ = Sum of all sides.

x + 2x - 5 + x + 6 = 29

4x + 1 = 29

4x = 29 - 1

4x = 28

x = 7.

Then, Other sides are

  • 1st sides = 2 × 7 - 5 = 9
  • 2nd sides = 7 + 6 = 13

Hence, The sides of triangle are 7, 9 & 13.

PolarNik [594]3 years ago
4 0

Answer:

  • 7, 9, 13

Step-by-step explanation:

Let the sides of the triangle be a, b and c.

<u>We have:</u>

  • P = a + b + c = 29
  • a = 2b - 5
  • c = b + 6

<u>Substitute values and solve for b:</u>

  • 2b - 5 + b + b + 6 = 29
  • 4b + 1 = 29
  • 4b = 28
  • b = 7

<u>Find a and c:</u>

  • a = 2*7 - 5 = 14 - 5 = 9
  • c = 7 + 6 = 13

<u>The sides are:</u> 7, 9, 13

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Choice B

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Choice C

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Answer:

The solutions to the quadratic function are:

x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}

Step-by-step explanation:

Given the function

f\left(x\right)\:=\:4x^2\:-\:4x\:+\:3

Let us determine all possible solutions for f(x) = 0

0=4x^2-4x+3

switch both sides

4x^2-4x+3=0

subtract 3 from both sides

4x^2-4x+3-3=0-3

simplify

4x^2-4x=-3

Divide both sides by 4

\frac{4x^2-4x}{4}=\frac{-3}{4}

x^2-x=-\frac{3}{4}

Add (-1/2)² to both sides

x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{3}{4}+\left(-\frac{1}{2}\right)^2

x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{1}{2}

\left(x-\frac{1}{2}\right)^2=-\frac{1}{2}

\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}

solving

x-\frac{1}{2}=\sqrt{-\frac{1}{2}}

x-\frac{1}{2}=\sqrt{-1}\sqrt{\frac{1}{2}}                 ∵ \sqrt{-\frac{1}{2}}=\sqrt{-1}\sqrt{\frac{1}{2}}

as

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x-\frac{1}{2}=i\sqrt{\frac{1}{2}}

Add 1/2 to both sides

x-\frac{1}{2}+\frac{1}{2}=i\sqrt{\frac{1}{2}}+\frac{1}{2}

x=i\sqrt{\frac{1}{2}}+\frac{1}{2}

also solving

x-\frac{1}{2}=-\sqrt{-\frac{1}{2}}

x-\frac{1}{2}=-i\sqrt{\frac{1}{2}}

Add 1/2 to both sides

x-\frac{1}{2}+\frac{1}{2}=-i\sqrt{\frac{1}{2}}+\frac{1}{2}

x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}

Therefore, the solutions to the quadratic function are:

x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}

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