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Vsevolod [243]
3 years ago
10

Please help me thank you ​

Mathematics
1 answer:
nirvana33 [79]3 years ago
3 0

Answer:

52 cm²

Step-by-step explanation:

There are 3 distinct faces:

one with sides 2 and 3

one with sides 2 and 4

one with sides 3 and 4

There are two of each distinct face (they are accros each other)

so we find their areas (we times by two as there are two of each face)

2×3×2=12

2×4×2=16

3×4×2=24

Finally we add the values:

12+16+24=52 cm²

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PLZ HELP, AND PLZ EXPLAIN
shutvik [7]

Answer:

C

Step-by-step explanation:

To make it easy let's start by organizing our information :

  • AC=12 AND BD=8
  • ABCD is a rhombus
  • K and L are the midpoints of sides AD and CD
  • we notice that the rhombus ABCD is divided into four right triangles

What do you think of when you hear a right triangle ?

  • The pythagorian theorem !

AC and BD  are khown so let's focus on them .

If we concentrated we can notice that AB and BD are cossing each other in the midpoints . why ?

Simply because they are the diagonals of a rhombus .

ow let's apply the pythagorian theorem :

  • (AC/2)² + (BD/2)² = BC²
  • 6²+4²=52
  • BC²= 52⇒\sqrt{52}=BC

Now we khow that : AB=BC=CD=AD=\sqrt{52}

This isn't enough . Let's try to figure out a way to calculate the length of KL  wich is the base of the triangle

  • KL is parallel to AC
  • k is the midpoint of AD and L of DC

I smell something . yes! Thales theorem

  • KL/AC=DL/DC=DK/AD WE4LL TAKE OLY ONE
  • KL/12=\sqrt{52}/2*\sqrt{52}  
  • KL/12=1/2⇒ KL=6

Now we have the length of the base kl

Now the big boss the height :

  • notice that you khow the length of KL
  • BD crosses kl from its midpoint and DL = \sqrt{52} /2

What I want to do is to apply the pythgorian thaorem to khow the lenght of that small part that is not a part of the height of the triangle . I will call it D

  • DL²=(KL/2)²+D²
  • 52/4= 9+ D²
  • D² = 52/4-9 +4 SO D=2

now the height of the trigle is H= BD-D= 8-2=6

NOw the area of the triangle is :

  • A=(KL*H)/2 ⇒ A= (6*6)/2=18

THE ANSWER IS 18 SQ.UN

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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21−=2(2−)=2cos(−1)+2 sin(−1)

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Is the above the correct way to write 21− and −1+2 in the form +? I wasn't sure if I could change Euler's formula to =cos()+sin(), where is a constant.

complex-numbers

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edited Mar 6 '17 at 4:38

Richard Ambler

1,52199 silver badges1616 bronze badges

asked Mar 6 '17 at 3:34

14wml

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Add a comment

1 Answer

1

No. It is not true that =cos()+sin(). Notice that

1=1≠cos()+sin(),

for example consider this at =0.

As a hint for figuring this out, notice that

+=ln(+)

then recall your rules for logarithms to get this to the form (+)ln().

8 0
3 years ago
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