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mel-nik [20]
3 years ago
11

in the given figure, AB//PQ//CD. AB = x units. CD = y units and PQ = z unites. Prove that 1/x + 1/y = 1/z

Mathematics
1 answer:
Arlecino [84]3 years ago
7 0
12. ∆ABD is similar to ∆PQD, so
   QD/z = BD/x
Likewise ∆CDB is similar to ∆PQB, so
   QB/z = BD/y

Since QD + QB = BD, you have
   QD/z + QB/z = BD/x + BD/y
   BD/z = BD/x + BD/y
Dividing by BD gives the desired result:
   1/x + 1/y = 1/z


13. Triangles ABD, BCD, and ACB are all similar. This means
   AB/AC = AD/AB
   AB² = AC×AD = (4 cm)×(9 cm)
   AB² = 36 cm²
   AB = 6 cm
and
   BD/CD = AD/BD
   BD² = CD×AD = (5 cm)×(4 cm)
   BD² = 20 cm²
   BD = 2√5 cm
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Answer:

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Step-by-step explanation:

y=mx+b

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3y=x+6,

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y=1/3x+2

slope is 1/3

8 0
3 years ago
A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuali
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Answer:

y                 0             1              2          3

P(Y=y)   0.0676   0.3549   0.3875   0.19

Step-by-step explanation:

P(Wed) = 0.26

P(Thurs) = 0.39

P(Fri) = 0.25

P(Sat) = 0.10

Y = No. of days beyond Wednesday it takes for both magazines to arrive i.e. 0,1,2,3

Y=0 means the magazines will arrive on Wednesday

Y=1 means the magazines will arrive till Thursday

Y=2 means the magazines will arrive till Friday

Y=3 means the magazines will arrive till Saturday

The possible combinations for Y are

Y(W,W) Y(W,T) Y(W,F) Y(W,S)

Y(T,W) Y(T,T) Y(T,F) Y(T,S)

Y(F,W) Y(F,T) Y(F,F) Y(F,S)

Y(S,W) Y(S,T) Y(S,F) Y(S,S)

So, we can classify these possible outcomes as Y=0,1,2,3.

Y(0) = Y(W,W) (both magazines take 0 days to arrive beyond Wednesday)

Y(1) = Y(W,T), Y(T,T), Y(T,W) (both magazines take 1 day to arrive beyond Wednesday)

Y(2) = Y(W,F), Y(T,F), Y(F,F) Y(F,W) Y(F,T) (both magazines arrive till Friday)

Y(3) = Y(W,S), Y(T,S), Y(F,S), Y(S,W), Y(S,T), Y(S,F), Y(S,S) (both magazines arrive till Saturday)

To calculate the PMF, we need to calculate the probability for each of the points in Y(0,1,2,3).

Y(0) = Y(W,W)

       = 0.26 x 0.26

Y(0) = 0.0676

Y(1) = Y(W,T) + Y(T,T) + Y(T,W)

      = (0.26 x 0.39) + (0.39 x 0.39) + (0.39 x 0.26)

      = 0.1014 + 0.1521 + 0.1014

Y(1) = 0.3549

Y(2) = Y(W,F) + Y(T,F) + Y(F,F) + Y(F,W) + Y(F,T)

  =(0.26 x 0.25) + (0.39 x 0.25) + (0.25 x 0.25) + (0.25 x 0.26) + (0.25 x 0.39)

  = 0.065 + 0.0975 + 0.0625 + 0.065 + 0.0975

Y(2) = 0.3875

Y(3) = Y(W,S) + Y(T,S) + Y(F,S) + Y(S,W) + Y(S,T) + Y(S,F) + Y(S,S)

      = (0.26 x 0.10) + (0.39 x 0.10) + (0.25 x 0.10) + (0.10 x 0.26) + (0.10 x 0.39) + (0.10 x 0.25) + (0.10 x 0.10)

       = 0.026 + 0.039 + 0.025 + 0.026 + 0.039 + 0.025 + 0.010

Y(3) = 0.19

y                 0             1              2          3

P(Y=y)   0.0676   0.3549   0.3875   0.19

The PMF plot is attached as a photo here.

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The sum of two consecutive integers is one less than three times the smaller integer. Find the two integers.
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The small number is 2.

The large number is 3.

<u>Step-by-step explanation:</u>

Let the two consecutive numbers be x and x+1.

  • x be the small integer.
  • x+1 be the large integer.

The sum of these two consecutive integers = small integer + large integer

The sum of these two consecutive integers is x+x+1 = (2x+1)

It is given that,

  • The sum of two consecutive integers is one less than three times the smaller integer.
  • This means that, (2x+1) is one less than three times the smaller integer.
  • Here, the small integer is represented as x.

<u>Therefore, it can determined that :</u>

(2x+1) = 3x-1

Keeping x term on one side and constants on other side,

3x-2x = 1+1

x = 2

Therefore, the small number is 2 and the large number is x+1 = 3.

3 0
3 years ago
Which of the following sets of numbers could be the lengths of the sides of a triangle?
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Answer:

  E.  2, 3, 4

Step-by-step explanation:

The sum of the shortest two sides must exceed the longest side. That is only the case for the set ...

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