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Leto [7]
3 years ago
7

Determine the correct set up for solving the equation using the quadratic formula.

Mathematics
1 answer:
netineya [11]3 years ago
5 0

Answer: B

Begin by simplifying the given equation, so it's rearranged into standard form. Move all the terms onto one side, so they equal 0. This would result with 4x^{2}-3x-9=0. Quadratic formula is (-b±√(b²-4ac))/(2a). Standard form follow this variable format: ax^{2}+bx+c. Use the numbers that correspond with each variable to substitute them into the quadratic formula. This would be

(-(-3)±√(-3²-4(4)(-9)/2(4).

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3 years ago
What is the value of (6a-2)
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Find the exact value of the indicated trigonometric functions for the acute angle a:
amid [387]

Tan=24/10 ≈ 2.4

using Pythagoras theorem

x²=(24)²+(10)²

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x²=57600

x=√57600 => 240

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7 0
3 years ago
H(x)=(x^2)+1 and K(x)=-(x^2)+4. If K(H)=0, what are the roots/solutions?
zubka84 [21]
H(x)=(x^2)+1\ \ \ \ and\ \ \ \  K(x)=-(x^2)+4. \\\\K(H)=-(H^2)+4=-(x^2+1)^2+4=4-(x^2+1)^2=\\\\.\ \ \ =2^2-(x^2+1)^2=(2-x^2-1)(2+x^2+1)=(1-x^2)(3+x^2)=\\\\.\ \ \ =(1-x)(1+x)(x^2-3\cdot i^2)=(1-x)(1+x)(x- \sqrt{3} \cdot i)(x+ \sqrt{3} \cdot i)\\\\ K(H)=0\ \ \ \ \Leftrightarrow\ \ \ (1-x)(1+x)(x- \sqrt{3} \cdot i)(x+ \sqrt{3} \cdot i)=0\\\\x=1\ \ \ \ or\ \ \ \ x=-1\ \ \ \ or\ \ \ \ x=\sqrt{3} \cdot i\ \ \ \ or\ \ \ \ x=-\sqrt{3} \cdot i\\\\Ans.\ e.
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3 years ago
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