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Crank
3 years ago
5

Solve. Good luck! Please do not try to google this.

Mathematics
1 answer:
Elena L [17]3 years ago
5 0
\frac{x^{2} + x - 2}{6x^{2} - 3x} = \sqrt{2x} + \frac{3x^{2}}{2}
\frac{x^{2} + 2x - x - 2}{3x(x) - 3x(1)} = \frac{2\sqrt{2x}}{2} + \frac{3x^{2}}{2}
\frac{x(x) + x(2) - 1(x) - 1(2)}{3x(x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}
\frac{x(x + 2) - 1(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}
\frac{(x - 1)(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}
\frac{(x - 1)(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}
3x(2x - 1)(2\sqrt{2x} + 3x^{2}) = 2(x + 2)(x - 1)
3x(2x(2\sqrt{2x} + 3x^{2}) - 1(2\sqrt{2x} + 3x^{2}) = 2(x(x - 1) + 2(x - 1))
3x(2x(2\sqrt{2x}) + 2x(3x^{2}) - 1(2\sqrt{2x}) - 1(3x^{2})) = 2(x(x) - x(1) + 2(x) - 2(1)
3x(4x\sqrt{2x} + 6x^{3} - 2\sqrt{2x} - 3x^{2}) = 2(x^{2} - x + 2x - 2)
3x(4x\sqrt{2x} - 2\sqrt{2x} + 6x^{3} - 3x^{2}) = 2(x^{2} + x - 2)
3x(4x\sqrt{2x}) - 3x(2\sqrt{2x}) + 3x(6x^{3}) - 3x(3x^{2}) = 2(x^{2}) + 2(x) - 2(2)
12x^{2}\sqrt{2x} - 6x\sqrt{2x} + 18x^{4} - 9x^{3} = 2x^{2} + 2x - 4
12x^{2}\sqrt{2x} - 6x\sqrt{2x} = -18x^{4} + 9x^{3} + 2x^{2} + 2x - 4
6x\sqrt{2x}(2x) - 6x\sqrt{2x}(1) = -9x^{3}(2x) - 9x^{3}(-1) + 2(x^{2}) + 2(x) - 2(2)
6x\sqrt{2x}(2x - 1) = -9x^{3}(2x - 1) + 2(x^{2} + x - 2)
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I Need Help Pwease :-&gt;<br> ******************************
MissTica

Answer:

Area: 135 ft^2

Perimeter: 50 ft

Step-by-step explanation:

area:

take the rectanle so length 12 x 9 = 108 so that is the length of the rectangle and now we need to find that of the triangle left over

subract 18 - 12 = 6 so that is the base of the trianle and we know the side length is 9 so plus it in A = (9)(6)/2

A = 54/2

A = 27

add 27 + 108 to get the total area

135

perimeter:

18 + 9 + 11 + 12 = 50

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\mathsf{sin\!\left(\dfrac{3\pi}{4}\right)}\\\\\\ =\mathsf{sin\!\left(\dfrac{3\cdot 180^\circ}{4}\right)}\\\\\\ =\mathsf{sin(135^\circ)}\\\\ =\mathsf{sin(180^\circ-135^\circ)}

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I hope this helps.=)


Tags:  <em>sine sin common angles trig trigonometry </em>

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