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The perimeters of similar triangles are in the same ratio as the corresponding sides

wariber [46]

Answer:

Always

Step-by-step explanation:

Suppose you have triangle ABC with side lengths a, b, c. Suppose that is similar to triangle DEF with side lengths d, e, f.

Now, let k be the ratio of corresponding sides ...

k = d/a

Because the same factor applies to all sides, we also have ...

k = e/b = f/c

That is, if we multiply by the denominators of each of these fractions, we get ...

d = a·k
e =b·k
f = c·k

The perimeter of ΔABC is ...

perimeter(ABC) = a + b + c

The perimeter of ΔDEF is ...

perimeter(DEF) = d + e + f = a·k + b·k + c·k

perimeter(DEF) = k(a + b + c) = k·perimeter(ABC)

k = perimeter(DEF)/perimeter(ABC)

That is, the perimeters are in the same ratio as corresponding sides.

3 0

3 years ago

Factor the polynomial function over the complex numbers. f(x)=x^4-x^3-2x−4 Enter your answer in the box. f(x) = The answer is: (

Vladimir79 [104]

Answer:

Factors: $(x+i\sqrt2)(x-i\sqrt2)(x+1)(x-2)=0$

Step-by-step explanation:

We are given a polynomial:

$f(x)=x^4-x^3-2x-4$

We have to factor the given polynomial into its complex factors.

The factorization can be done as follows:

$f(x)=x^4-x^3-2x-4 = 0\\x^4-x^3-2x-4 = 0\\x^4-4-x^3-2x=0\\\text{Identity: }a^2-b^2 = (a+b)(a-b)\\(x^4-4)-(x^3+2x) = 0\\(x^2+2)(x^2-2)-x(x^2+2) = 0\\(x^2+2)(x^2-2-x) = 0\\(x^2+2)(x^2-x-2) = 0\\(x^2+2)(x^2-2x-+x-2) = 0\\(x^2+2)((x(x-2)+1(x-2))=0\\(x^2+2)(x+1)(x-2)=0\\\text{Identity: }a^2-b^2 = (a+b)(a-b)\\(x^2-(\sqrt{-2})^2)(x+1)(x-2)=0\\(x+i\sqrt2)(x-i\sqrt2)(x+1)(x-2)=0$