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vazorg [7]
2 years ago
8

Which of the following points lies on the circle whose center is at the origin and whose radius is 10?

Mathematics
1 answer:
Romashka-Z-Leto [24]2 years ago
6 0

The equation is: \frac{(x - 0)^{2} }{10^{2} } + \frac{(y - 0)^{2} }{10^{2} } = 1

= \frac{x^{2} }{10^{2} } + \frac{y^{2} }{10^{2} } = 1

A) \frac{(\sqrt{10}) ^{2} }{10^{2} } + \frac{(0)^{2} }{10^{2} } = 1

⇒ \frac{10}{100} + 0 = 1   FALSE

B) \frac{(0) ^{2} }{10^{2} } + \frac{(2\sqrt{5})^{2} }{10^{2} } = 1

⇒ 0 + \frac{20}{100} = 1   FALSE

C) \frac{(5\sqrt{2}) ^{2} }{10^{2} } + \frac{(5\sqrt{2})^{2} }{10^{2} } = 1

⇒ \frac{50}{100} + \frac{50}{100} = 1   TRUE

Answer: C


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Drag each tile to the correct box consider the given functions f,g and h place the tiles in order from least to greatest accordi
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Answer:

Step-by-step explanation:

By definition, the average rate of change of a function f over an interval [a,b] is given by

\dfrac{f(b)-f(a)}{b-a}

compute the quantity

\dfrac{f(3)-f(0)}{3}

for all the three function

Average rate of change of f:

We will simply use the table to check the values for f(3) and f(0):

\dfrac{f(3)-f(0)}{3}=\dfrac{10-1}{3} = 3

Average rate of change of g:

We will use the graph to to check the values for g(3) and g(0):

\dfrac{g(3)-g(0)}{3}=\dfrac{8-1}{3} = \dfrac{7}{3}

Average rate of change of h:

We can plug the values in the equation to get h(3) and h(0):

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And so the average rate of change is

\dfrac{h(3)-h(0)}{3}=\dfrac{6-(-6)}{3} = 4

3 0
3 years ago
Explain how you have solved for x and show your workings out. I will award the brainliest 25 points!
Brums [2.3K]

Answer:

The value of x is 10º.

Step-by-step explanation:

From Euclidean Geometry we remember that the sum of internal angles within a triangle equals to 180º. We present the resulting triangle after applying some geometric handling in the image attached below. Then, the triangle satisfies the following equation:

100^{\circ} + (80^{\circ}-5\cdot x)+(4\cdot x +10^{\circ}) = 180^{\circ} (1)

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The value of x is 10º.

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Answer:

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6 0
3 years ago
Can someone please help me factor this
Dmitry_Shevchenko [17]

Answer:

\huge\boxed{\bf\:1}

Step-by-step explanation:

\frac{ x  ^ { 2  }  -4x+3  }{ x  ^ { 2  }  -7x+12  }   \times   \frac{ x  ^ { 2  }  +2x-24  }{ x  ^ { 2  }  +5x-6  }    ^ {   }

Take \frac{ x  ^ { 2  }  -4x+3  }{ x  ^ { 2  }  -7x+12  } & factorise it at first.

\frac{ x  ^ { 2  }  -4x+3  }{ x  ^ { 2  }  -7x+12  }   \\= \frac{\left(x-3\right)\left(x-1\right)}{\left(x-4\right)\left(x-3\right)}\\= \frac{x-1}{x-4}

Now factorise the next set : \frac{ x  ^ { 2  }  +2x-24  }{ x  ^ { 2  }  +5x-6  }    ^ {   }.

\frac{ x  ^ { 2  }  +2x-24  }{ x  ^ { 2  }  +5x-6  }    ^ {   }\\= \frac{\left(x-4\right)\left(x+6\right)}{\left(x-1\right)\left(x+6\right)}\\=  \frac{x-4}{x-1}

Now, multiply the two simplified results.

\frac{ x  ^ { 2  }  -4x+3  }{ x  ^ { 2  }  -7x+12  }   \times   \frac{ x  ^ { 2  }  +2x-24  }{ x  ^ { 2  }  +5x-6  }    ^ {   }\\= \frac{x-1}{x-4}\times \frac{x-4}{x-1} \\= \frac{\left(x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x-1\right)} \\= \boxed{\bf\: 1}

\rule{150pt}{2pt}

7 0
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