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Licemer1 [7]
3 years ago
15

How does the function f(x) = a ln x compare to the parent function when |a| > 1?

Mathematics
1 answer:
VladimirAG [237]3 years ago
5 0

Answer: The graph is stretched.

Step-by-step explanation:

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A graph is a function if there is only one X value for any Y value.

Since the graph is in the shape of a U, there can only be one X for every Y value, so this graph is a function.
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Solve the equation graphically: 12x-7y=-6 5x-7y=27
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SVETLANKA909090 [29]

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x² + 3x - 28

Step-by-step explanation:

Each term in the second factor is multiplied by each term in the first factor

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Please help! 16/20. Will give brainliest! Spam answers will not be tolerated!
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PLEASE HELP!!
erik [133]
2.=c.\\\\2\sin4x\cos4x=2\sin(2\cdot4x)=2\sin8x\\\\Used:\\\sin2\alpha=2\sin\alpha\cos\alpha

1.=b.\\\\\csc x-\sin x=\dfrac{1}{\sin x}-\dfrac{\sin^2x}{\sin x}=\dfrac{1-\sin^2x}{\sin x}=\dfrac{\cos^2x}{\sin x}\\\\=\dfrac{\cos x\cos x}{\sin x}=\cos x\cdot\dfrac{\cos x}{\sin x}=\cos x\cot x\\\\Used:\\\csc x=\dfrac{1}{\sin x}\\\\\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\\cot x=\dfrac{\cos x}{\sin x}

3.=a.\\\\\dfrac{\sin x-1}{\sin x+1}=\dfrac{\sin x-1}{\sin x+1}\cdot\dfrac{\sin x+1}{\sin x+1}=\dfrac{\sin^2x-1^2}{(\sin x+1)^2}=\dfrac{\sin^2x-1}{(\sin x+1)^2}\\\\=\dfrac{-(1-\sin^2x)}{(\sin x+1)^2}=\dfrac{-\cos^2x}{(\sin x+1)^2}\\\\Used:\\(a-b)(a+b)=a^2-b^2\\\\\sin^2x+\cos^2x=1\to \cos^2x=1-\sin^2x
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3 years ago
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