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boyakko [2]
3 years ago
7

Given sec (theta) = 3 what is the exact value of csc (90 - (theta))

Mathematics
1 answer:
sp2606 [1]3 years ago
5 0
Sec θ = 3
1/cos θ = 3
cos θ = 1/3

csc (90 - θ) = 1/sin (90 - θ) = 1/cos θ = 1/(1/3) = 3

Therefore, csc (90 - θ) = 3.
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Cscx/secx = cotx<br><br> Verify with steps please :)
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cot(x) = cot(x)
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3 years ago
Solve.
Ksenya-84 [330]
<span>Solve.

x² + 4x + 4 = 18


​ A x=−4±3√ ​2

​ B x=2±3√ ​2

​ C x=4±9√ ​2

D x=−2±3√2



</span>x^2 + 4x + 4 = 18 \\  \\ x^2+4x+4-18= 0 \\  \\ x^2+4x-14=0 \\  \\ x_1_y_2=  \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} \qquad a= 1\qquad b= 4\qquad c= -14 \\  \\  \\  x_1_y_2=  \dfrac{-4\pm \sqrt{4^2-4(1)(-14)} }{2(1)} \\  \\  \\  x_1_y_2=  \dfrac{-4\pm \sqrt{16-(-56)} }{2}  \\  \\  \\  x_1_y_2=  \dfrac{-4\pm \sqrt{72} }{2}  \\  \\  \\  x_1=  \dfrac{-4+ \sqrt{72} }{2} \qquad\qquad x_2=  \dfrac{-4+ \sqrt{72} }{2}\\  \\  \\  x_1=  \dfrac{-4+ \sqrt{2^3*3^2} }{2} \qquad\qquad x_2=  \dfrac{-4- \sqrt{2^3*3^2} }{2}
<span>
</span>x_1=  \dfrac{-4+2*3 \sqrt{2} }{2} \qquad\qquad x_2=  \dfrac{-4- 2*3\sqrt{2} }{2} \\  \\  \\ x_1=  \dfrac{-4+6 \sqrt{2} }{2} \qquad\qquad x_2=  \dfrac{-4- 6\sqrt{2} }{2} \\  \\  \\ x_1=  \dfrac{-4}{2} + \dfrac{6 \sqrt{2} }{2} \qquad\qquad x_2=   \dfrac{-4}{2}- \dfrac{ 6\sqrt{2} }{2} \\  \\  \\  x_1= -2 + 3 \sqrt{2} \qquad\qquad\quad  x_2=  -2- 3\sqrt{2}  \\  \\  \\ \boxed{x=   -2\pm 3\sqrt{2} }  \to D)<span>
</span>
7 0
3 years ago
Read 2 more answers
Write an expression for the number of yards in f feet
koban [17]
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f = 3y
3 0
3 years ago
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