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dmitriy555 [2]
4 years ago
11

The ratios of corresponding sides in the two triangles are equal.

Mathematics
2 answers:
olga55 [171]4 years ago
7 0

Answer:

∠I ≅ ∠F

Step-by-step explanation:

Since, F is common vertex in the sides GF and FE of triangle FGE and I is common vertex in the sides IJ and IH of triangle IJH.

Hence, ∠I ≅ ∠F will be the required information to prove that △FGE ~ △IJH by the SAS similarity theorem.

Alisiya [41]4 years ago
6 0

Answer:

∠I ≅ ∠F

Step-by-step explanation:

I took the test.

Got it right ;3

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10/28

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Step-by-step explanation:

5/14

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3 years ago
How do I determine z ∈ C:
saw5 [17]

Simplify the coefficient of z on the left side. We do this by rationalizing the denominators and multiplying them by their complex conjugates:

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3-2i}{1+i}\cdot\dfrac{1-i}{1-i} - \dfrac{5+3i}{1+2i}\cdot\dfrac{1-2i}{1-2i}

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{(3-2i)(1-i)}{1-i^2} - \dfrac{(5+3i)(1-2i)}{1-(2i)^2}

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3 - 2i - 3i + 2i^2}{1-(-1)} - \dfrac{5 + 3i - 10i - 6i^2}{1-4(-1)}

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3 - 5i + 2(-1)}2 - \dfrac{5 - 7i - 6(-1)}5

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{1 - 5i}2 - \dfrac{11 - 7i}5

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{1 - 5i}2\cdot\dfrac55 - \dfrac{11 - 7i}5\cdot\dfrac22

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{5 - 25i - 22 + 14i}{10}

\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = -\dfrac{17 + 11i}{10}

So, the equation is simplified to

-\dfrac{17+11i}{10} z = \dfrac12 - \dfrac{2i}5

Let's combine the fractions on the right side:

\dfrac12 - \dfrac{2i}5 = \dfrac12\cdot\dfrac55 - \dfrac{2i}5\cdot\dfrac22

\dfrac12 - \dfrac{2i}5 = \dfrac{5-4i}{10}

Then

-\dfrac{17+11i}{10} z = \dfrac{5-4i}{10}

reduces to

-(17+11i) z = 5-4i

Multiply both sides by -1/(17 + 11i) :

\dfrac{-(17+11i)}{-(17+11i)} z = \dfrac{5-4i}{-(17+11i)}

z = -\dfrac{5-4i}{17+11i}

Finally, simplify the right side:

-\dfrac{5-4i}{17+11i} = -\dfrac{5-4i}{17+11i} \cdot \dfrac{17-11i}{17-11i}

-\dfrac{5-4i}{17+11i} = -\dfrac{(5-4i)(17-11i)}{17^2-(11i)^2}

-\dfrac{5-4i}{17+11i} = -\dfrac{85 - 68i - 55i + 44i^2}{289-121(-1)}

-\dfrac{5-4i}{17+11i} = -\dfrac{85 - 68i - 55i + 44(-1)}{410}

-\dfrac{5-4i}{17+11i} = -\dfrac{41 - 123i}{410}

-\dfrac{5-4i}{17+11i} = -\dfrac{41 - 41\cdot3i}{410}

-\dfrac{5-4i}{17+11i} = -\dfrac{1 - 3i}{10}

So, the solution to the equation is

z = -\dfrac{1-3i}{10} = \boxed{-\dfrac1{10} + \dfrac3{10}i}

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hi,

You have to use a formula which is  known as of binomial coefficient which is :

k among n  is  :   C ( n /k) =   n!  / k! ( n-k)!

! = factorial

ex :   5 object among 8 is :    C ( 8/5 ) = 8 ! / 5! * ( 8-5) !

As 8! =  8*7*6*5*4*3*2*1 =  40 320

As 5! = 5*4*3*2*1 = 120  

as 3 ! = 3*2*1 = 6

then :     40 320  / 120 * 6

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Conclusion : there is  56 way to select  5 object among 8 of them.  

ps : you can also use Pascal Triangle to do such a thing.

               

3 0
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