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Olin [163]
2 years ago
5

At a taffy pull, George stretched the taffy to 3 feet. Jose stretched it 1_ 1 3 times as far as George. Maria stretched it _ 2 3

as far as George. Sally stretched it _ 6 6 as far. Who stretched it the farthest? The least?
Mathematics
1 answer:
alexira [117]2 years ago
6 0

Answer:

Jose stretched the taffy the farthest and Maria stretched it the least

Step-by-step explanation:

Distance stretched by George = 3 feet

Distance stretched by Jose is 1\dfrac{1}{3} times as much as George so

3\times 1\dfrac{1}{3}=3\times\dfrac{4}{3}=4\ \text{feet}

Distance stretched by Jose is 4\ \text{feet}

Maria streched the taffy \dfrac{2}{3} times as George

3\times \dfrac{2}{3}=2\ \text{feet}

Maria stretched the taffy by 2\ \text{feet}

Sally stretched it \dfrac{6}{6} so 1\ \text{feet}

So, Jose stretched the taffy the farthest and Maria stretched it the least.

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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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Answer:

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Step-by-step explanation:

make the equation

x+x+2+x+4+x+6=72

simplify

4x+12=72

x+3=18

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