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Dafna11 [192]
3 years ago
7

A square garden has an area of 441 m^2.if the path of uniform width surrounding inside the garden is 216m^2,find the width of th

e path
Mathematics
1 answer:
Ilya [14]3 years ago
4 0

Answer:

Area of a square garden = 2, 500 m².

Length of the side of the garden =  √2,500 = 50 m.

Width of the path = 2.5 m.

Path is inside the garden.

Width of the garden that is inside (enclosed) by the path:

         50 - 2 * 2.5 = 45 m.

Length of the garden inside by the path :  = 45 m.

So the area of garden inside the path = 45 * 45 m²

      = 2, 025 m².

Area of the path = 2, 500 - 2, 025 

     = 475 m².

Cost of laying the path = Rs 25 /m² * 475 m²

    = Rs 11, 875.

<u>MARK THIS AS BRAINLIEST</u>

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Moes sells 10 tacos for $8.49 or Taco Bell sells 6 of the same kind of taco for $5.40. Which is a better deal? Clearly show the
Degger [83]

Answer:

The one that sells 10 tacos for $8.49

To get the better deal, you should find the selling prices for a single tacos in both cases  and compare the amounts.

In the first one, 10 tacos for $8.49

This means 1 tacos sells for $8.49/10 = $0.849

In the second one, 6 for $5.40, this means for a single tacos, the selling price is $5.40/6 =$0.9

The better deal is the first one that sells one taco for $0.849 compared to the second one that sells one tacos for $0.9

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the answer is 7x, becuase -1 + 8 is 7


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4 years ago
Which expression is a fourth root of -1+isqrt3?
aleksklad [387]

Answer:

Step-by-step explanation:

\sf n^{th} roots of a complex number is given by DeMoivre's formula.

   \sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}

Here, k lies between 0 and (n -1) ; n is the exponent.

\sf -1 + i\sqrt{3}

a = -1 and b = √3

\sf \boxed{r=\sqrt{a^2+b^2}} \ and \ \boxed{\theta = Tan^{-1} \ \dfrac{b}{a}}

\sf r = \sqrt{(-1)^2 + 3^2}\\\\ = \sqrt{1+9}\\\\=\sqrt{10}

                   \sf \theta = tan^{-1} \ \dfrac{\sqrt{3}}{-1}\\\\ = tan^{-1} \ (-\sqrt{3})

                   \sf = \dfrac{-\pi }{3}

n = 4

For k = 0,

          \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin  \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi  }{12}+iSin  \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]

For k =1,

         \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{5\pi}{12}+i \ Sin \ \dfrac{5\pi}{12}\right]

For k =2,

       z = \sqrt[4]{10}\left[Cos \ \dfrac{11\pi}{12}+i \ Sin \ \dfrac{11\pi}{12}\right]

For k = 3,

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For k = 4,

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2 years ago
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BlackZzzverrR [31]

Answer:

Ok, here goes:  f(3) means take f(x) and plug in 3 wherever we see x.  So

f(3) = 12*39 + 3 = 236196 + 3

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Likewise, f(-3) means take f(x) and plug in (-3) wherever we see x.  So

f(-3) = 12*(-3)9 + (-3) = -236196 - 3

f(-3) = -236199

Putting those two things together, we get

f(3) + f(-3) = 236199 - 236199 = 0

Step-by-step explanation:

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