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Anastaziya [24]
3 years ago
8

A shopkeeper of Banepa bought 300 pieces of same type of the sari. He sold

Mathematics
1 answer:
Leviafan [203]3 years ago
4 0

Answer:

Rs. 1315

Step-by-step explanation:

He bought 300 saris

0.7(300) were sold at 1500 / piece

0.3(300) were sold at 1350/ piece

210 were sold at 1500/ piece = 315000

90 were sold at 1350/piece = 121500

Total amount  = 315000+121500 = 436500

Profit = 42000

Subtract:

436500 - 42000 = 394500

394500 is the price for 300 saris

Divide by 300

394500 / 300 = 1315

He purchased one sari at 1315

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

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Find each lettered angel measure
hammer [34]

Answer:

a = 64°

b = 138.7°

Step-by-step explanation:

First, Let's find the measure a,

  • 180° = 116° + a
  • 180 - 116 = a
  • a = 64°

______

Now,

  • 90° + a + 82° + x + x + x = 360°
  • 90+ 64 + 82 + 3x = 360°
  • 236 + 3x = 360
  • 3x = 124/ 3
  • x = 41.3°

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Now that we found out what x equals, let's find 'b'-

  • b + x = 180°
  • b + 41.3 = 180
  • b = 180 - 41.3
  • b = 138.7°
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Is x plus y equals 3, then 3 equals x plus y. what is the property
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How many solutions does the system y=6x-4 and y=4-6x have?
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3 years ago
Find the point P on the graph of the function y=√x closest to the point (9,0)
Sphinxa [80]

Answer:

\displaystyle \frac{17}{2}.

Step-by-step explanation:

Let the x-coordinate of P be t. For P\! to be on the graph of the function y = \sqrt{x}, the y-coordinate of \! P would need to be \sqrt{t}. Therefore, the coordinate of P \! would be \left(t,\, \sqrt{t}\right).

The Euclidean Distance between \left(t,\, \sqrt{t}\right) and (9,\, 0) is:

\begin{aligned} & d\left(\left(t,\, \sqrt{t}\right),\, (9,\, 0)\right) \\ &= \sqrt{(t - 9)^2 +\left(\sqrt{t}\right)^{2}} \\ &= \sqrt{t^2 - 18\, t + 81 + t} \\ &= \sqrt{t^2 - 17 \, t + 81}\end{aligned}.

The goal is to find the a t that minimizes this distance. However, \sqrt{t^2 - 17 \, t + 81} is non-negative for all real t\!. Hence, the \! t that minimizes the square of this expression, \left(t^2 - 17 \, t + 81\right), would also minimize \sqrt{t^2 - 17 \, t + 81}\!.

Differentiate \left(t^2 - 17 \, t + 81\right) with respect to t:

\displaystyle \frac{d}{dt}\left[t^2 - 17 \, t + 81\right] = 2\, t - 17.

\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2.

Set the first derivative, (2\, t - 17), to 0 and solve for t:

2\, t - 17 = 0.

\displaystyle t = \frac{17}{2}.

Notice that the second derivative is greater than 0 for this t. Hence, \displaystyle t = \frac{17}{2} would indeed minimize \left(t^2 - 17 \, t + 81\right). This t\! value would also minimize \sqrt{t^2 - 17 \, t + 81}\!, the distance between P \left(t,\, \sqrt{t}\right) and (9,\, 0).

Therefore, the point P would be closest to (9,\, 0) when the x-coordinate of P\! is \displaystyle \frac{17}{2}.

8 0
3 years ago
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