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Zanzabum
4 years ago
11

A triangle has side lengths of (1.2u+1.8v)(1.2u+1.8v) centimeters, (6.5u+8.7w)(6.5u+8.7w) centimeters, and (5.8w-6.9v)(5.8w−6.9v

) centimeters. Which expression represents the perimeter, in centimeters, of the triangle? 13.5uw+3.6vw13.5uw+3.6vw 7.7u+1.8w+7.6v7.7u+1.8w+7.6v -1.1vw+15.2uw+3uv−1.1vw+15.2uw+3uv 14.5w-5.1v+7.7u14.5w−5.1v+7.7u
Mathematics
1 answer:
Komok [63]4 years ago
5 0

Answer:

7.7u - 5.1v + 14.5w

Step-by-step explanation:

Given length:

Side A = (1.2u+1.8v) centimeters,

Side B = (6.5u+8.7w) centimeters

Side C = (5.8w-6.9v) centimeters

Perimeter of a triangle = side A + side B + side C

= (1.2u+1.8v) + (6.5u+8.7w) + (5.8w-6.9v)

= 1.2u + 1.8v + 6.5u + 8.7w + 5.8w - 6.9v

Collect like terms

= 1.2u + 6.5u + 1.8v - 6.9v + 8.7w + 5.8w

= 7.7u - 5.1v + 14.5w

Therefore, the perimeter of the triangle is 7.7u - 5.1v + 14.5w

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Step-by-step explanation:

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3 0
2 years ago
(4x – 8)(3x + 9) = 0​
Gwar [14]

Answer:

\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

x=2,\:x=-3

Step-by-step explanation:

Considering the equation

                         \left(4x\:-\:8\right)\left(3x\:+\:9\right)\:=\:0

solving the equation

\left(4x\:-\:8\right)\left(3x\:+\:9\right)\:=\:0

\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

\mathrm{Solve\:}\:4x-8=0

4x-8=0

4x-8+8=0+8

4x=8

x=2

\mathrm{Solve\:}\:3x+9=0

3x+9=0

3x+9-9=0-9

3x=-9

x=-3

\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

x=2,\:x=-3

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4 years ago
Is the following expression shown the exact answer? 144π - 216√3<br><br> true or false
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6 0
4 years ago
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Aleksandr [31]

Answer:

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Step-by-step explanation:

5 0
3 years ago
Use the Remainder Theorem to determine which of the roots are roots of F(x). Show your work.
Brums [2.3K]

Answer:    x1=1   x2=-2  and x3=2

Step-by-step explanation:

1st   x1=1 is 1 of the roots , so

F(1)=1-1-4+4=0 - true

So lets divide x^3-x^2-4x+4 by (x-x1), i.e  (x^3-x^2-4x+4) /(x-1)=(x^2-4)

x^2-4 can be factorized as (x-2)*(x+2)

So x^3-x^2-4x+4=(x-1)*(x^2-4)=(x-1)(x-2)*(x+2)

So there are 3 dofferent roots:

x1=1   x2=-2  and x3=2

3 0
3 years ago
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