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ycow [4]
3 years ago
7

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

Mathematics
1 answer:
wariber [46]3 years ago
3 0

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.

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3 years ago
Please answer a and b
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IRISSAK [1]

Answer:

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Step-by-step explanation:

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3 years ago
Suppose that P(A) = 0.25 and P(B) = 0.40 . If P(A|B)=0.20 , what is P(B|A) ?
Ket [755]

Answer:

  • 0.32

Step-by-step explanation:

<u>Use of formula:</u>

  • P(A and B) = P(A)*P(B|A) and
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<u>According to above and based on given:</u>

  • P(A)P(B|A) = P(B)P(A|B)
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8 0
3 years ago
Write an equation for the nth term of the geometric sequence. Then find the 7th term. 2,-6,18,-54,...
Sonja [21]

soln,

from the series,

a = 2

t2 = -6

r = t2/a

=   \frac{ - 6}{2}

= -3

therefore, tn =

{ar}^{n - 1}

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