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d1i1m1o1n [39]
3 years ago
14

Using the quadratic formula to solve 4x^2-3x + 9=2x + 1 what are the values of x​

Mathematics
1 answer:
Alchen [17]3 years ago
8 0

Answer:

x=5±√−103/8

Step-by-step explanation:

There's no real solutions

4x2−3x+9−(2x+1)=2x+1−(2x+1)

4x2−5x+8=0

x=−b±√b2−4ac/2a

x=−(−5)±√(−5)2−4(4)(8)/2(4)

x=5±√−103/8

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If sin ⁡x=725, and 0 ∠ x ∠ pi/2, what is the tan (x - pi/4)
krok68 [10]

Tan(x-\frac{\pi }{4}) = \frac{-17}{31}

<u>Step-by-step explanation:</u>

Here we have ,  sin ⁡x=7/25( given sin x = 725 which is not possible ) , 0 . Let's find tan (x - pi/4):

⇒ Tanx = \frac{sinx}{cosx}

⇒ Tanx = \frac{sinx}{\sqrt{1-(sinx)^{2}}}

⇒ Tanx = \frac{\frac{7}{25}}{\sqrt{1-(\frac{7}{25})^{2}}}

⇒ Tanx = {\frac{7}{25}}{\sqrt{(\frac{625}{625-49})^{}}}

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Now , Tan(x-\frac{\pi }{4}) = \frac{Tanx - Tan(\frac{\pi }{4} )}{1+ Tanx(Tan(\frac{\pi }{4} )}

⇒ Tan(x-\frac{\pi }{4}) = \frac{Tanx -1}{1+ Tanx(1)}

⇒ Tan(x-\frac{\pi }{4}) = \frac{\frac{7}{24}  -1}{1+\frac{7}{24} }

⇒ Tan(x-\frac{\pi }{4}) = \frac{\frac{7-24}{24} }{\frac{7+24}{24} }

⇒ Tan(x-\frac{\pi }{4}) = \frac{-17}{24} (\frac{24}{31} )

⇒ Tan(x-\frac{\pi }{4}) = \frac{-17}{31}

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Step-by-step explanation:

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