The equation for the line of reflection is x = 2.
If you drew a vertical line crossing the x-axis on x = 2, the drawing will be the same on the left and right side, as if one side was drawn then exactly copied on the other side but flipped.
The figure is symmetrical, and x = 2 is the line of symmetry.
Hope this helps!

8 0

2 years ago

Helpppppppppppppppppppppppppppppp

Diano4ka-milaya [45]

In this case it is to find the roots of the polynomial.
We have then:
2x ^ 2-5x + 1 = 3
Rewriting:
2x ^ 2-5x-2 = 0
Applying resolver we have
x = (- b +/- root (b ^ 2 - 4ac)) / (2a)
Substituting values:
x = (- (- 5) +/- root ((- 5) ^ 2 - 4 (2) (- 2))) / (2 (2))
x = (- (- 5) +/- root ((25 + 16)) / (2 (2))
x = (5 +/- root (41))) / (4)
x = ((5/4) +/- (root (41)) / 4)
Answer:
x = ((5/4) +/- (root (41)) / 4)
(option 4)

8 0

3 years ago

Anyone know how to solve this??

Tom [10]

$Given : \frac{(x^\frac{2}{5})^9\;.\;(x^\frac{-4}{15})}{x^\frac{1}{3}}$

$\implies {(x^\frac{18}{5})\;.\; (x^\frac{-4}{15})}\;.\;(x^\frac{-1}{3})$

$\implies x^(^\frac{18}{5} ^-^\frac{4}{15}^-^\frac{1}{3}^)$

$\implies x^(^(^\frac{54 - 4}{15}^)^-^\frac{1}{3}^) = x^(^(^\frac{50}{15}^)^-^\frac{1}{3}^) = x^(^(^\frac{10}{3}^)^-^\frac{1}{3}^) = x^(^\frac{9}{3}^) = x^3$

$\implies x^k = x^3$

$\implies k = 3$

7 0

3 years ago