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makvit [3.9K]
3 years ago
15

What is the probability that a point chosen at random in the given figure will be inside the larger square and outside the small

er square? Enter your answer, as a fraction in simplest form, in the box. P(inside larger square and outside smaller square) = $\text{Basic}$ $x$$y$$x^2$$\sqrt{ }$$\frac{x}{ }$ $x\frac{ }{ }$ $x^{ }$$x_{ }$$\degree$$\left(\right)$$\abs{ }$$\pi$$\infty$ A square with a side length of 7 centimeters is within a larger square with a side length of 10 centimeters. The sides of the inner square do not touch the sides of the outer square.
Mathematics
1 answer:
liraira [26]3 years ago
5 0

Answer:

  • 51/100

Step-by-step explanation:

<em>The probability that a point chosen at random in the given figure will be inside the larger square and outside the smaller square</em> is equal to the ratio of the area of interest to the total area:

<em>P(inside larger square and outside smaller square)</em> = area of interest / total area

<em>P(inside larger square and outside smaller square)</em> = area inside the larger square and outside the smaller square / area of the larger square

<u>Calculations:</u>

<u />

1. <u>Area inside the larger square</u>: side² = (10 cm)² = 100 cm²

2. <u>Area inside the smaller square </u>= side² = (7cm)² = 49 cm²

3. <u>Area inside the larger square and outside the smaller square</u>

  • 100 cm² - 49 cm² = 51 cm²

4.<u> P (inside larger square and outside smaller squere)</u>

  • 51 cm² / 100 cm² = 51/100
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Determine whether the three points are the vertices of a right triangle. (6, -6), (9, -6), (9, 1)
lisabon 2012 [21]

Answer:

The three given points are the vertices of a right triangle.

Step-by-step explanation:

To determine that the three points are the vertices of a right triangle let us find the distance between each two points

The formula of the distance is d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }

∵ x_{1} = 6 and y_{1} = -6

∵ x_{2} = 9 and y_{2} = -6

∴ d_{1} = \sqrt{(9-6)^{2}+(-6--6)^{2}}=\sqrt{9+0}

∴ d_{1} = 3

∵ x_{1} = 6 and y_{1} = -6

∵ x_{2} = 9 and y_{2} = 1

∴ d_{2} = \sqrt{(9-6)^{2}+(1--6)^{2}}=\sqrt{9+49}

∴ d_{2} = \sqrt{58}

∵ x_{1} = 9 and y_{1} = 1

∵ x_{2} = 9 and y_{2} = -6

∴ d_{3} = \sqrt{(9-9)^{2}+(-6-1)^{2}}=\sqrt{0+49}

∴ d_{3} = 7

<em>Let us use the fact that if the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle at the vertex which opposite to the longest side</em>

∵ The longest side is \sqrt{58}

∵ The other two sides are 3 and 7

∵ ( \sqrt{58} )² = 58

∵ (3)² + (7)² = 9 + 49 = 58

∴  ( \sqrt{58} )² = (3)² + (7)²

- By using the fact above

∴ The triangle is a right triangle

The three given points are the vertices of a right triangle.

8 0
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Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{13}}}}}

Step-by-step explanation:

Given, f = 3 and g = 5

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Plug the values

⇒\sf{3 +  {5}^{2}  -  3 \times 5}

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⇒\sf{3 + 25 - 3 \times 5}

Multiply the numbers : 3 and 5

⇒\sf{3 + 25 - 15}

Add the numbers : 3 and 25

⇒\sf{28 - 15}

Subtract 15 from 28

⇒\sf{13}

Hope I helped!

Best regards! :D

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Step-by-step explanation:

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