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velikii [3]
3 years ago
8

PLEASE SOLVE FOR ME!!!! I am stuck on question... 36/5x - 10 < 5

Mathematics
1 answer:
FromTheMoon [43]3 years ago
3 0
*\   \dfrac{36}{5x} - 10 \ \textless \ 5 \qquad x \neq 0 \\  \\  \\   \dfrac{36}{5x}\ \textless \ 5 + 10  \\  \\  \\   \dfrac{36}{5x}\ \textless \ 15 \\  \\  \\  36\ \textless \  15 * (5x)  \\  \\ 36 \ \textless \  75x  \\  \\  \dfrac{36}{75}\ \textless \  x  \qquad    \dfrac{12}{25}\ \textless \  x  \qquad   \boxed{x\in ( \frac{12}{25}; +\infty)  }



 *\ \dfrac{36}{5x-10}  \ \textless \ 5 \qquad x \neq 2 \\ \\ \\36 \textless \ 5 (5x-10)  \\ \\ 36 \textless \ 25x-50 \\ \\ 36+50\ \textless \25x \\ \\ 86 \ \textless \ 25x \\ \\ \dfrac{86}{25}\ \textless \ x\qquad \to  \qquad \boxed{x\in ( \frac{86}{25}; +\infty) }
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Answer:

  28

Step-by-step explanation:

The attached figure shows 7 triangles (counting the original 2). Each has degree-4 rotational symmetry about the center of the figure, so there can be 7×4 = 28 triangles.

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Yuliya22 [10]

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125

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5 0
3 years ago
The standard form of the equation for a quadratic function is, with he vertex of the graph of the function at the point (p,q). t
ludmilkaskok [199]

Answer:

p = ½ (x₁ + x₂)

q = a (x₁x₂ − ¼ (x₁ + x₂)²)

Step-by-step explanation:

y = a (x − x₁) (x − x₂)

Expand:

y = a (x² − x₁x − x₂x + x₁x₂)

y = a (x² − (x₁ + x₂)x + x₁x₂)

Distribute a to the first two terms:

y = a (x² − (x₁ + x₂)x) + ax₁x₂

Complete the square:

y = a (x² − (x₁ + x₂)x + ¼(x₁ + x₂)²) + ax₁x₂ − ¼ a(x₁ + x₂)²

y = a (x − ½ (x₁ + x₂))² + a (x₁x₂ − ¼ (x₁ + x₂)²)

Therefore:

p = ½ (x₁ + x₂)

q = a (x₁x₂ − ¼ (x₁ + x₂)²)

3 0
3 years ago
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