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nikdorinn [45]
2 years ago
14

Help me dawg. I need this

Mathematics
1 answer:
prohojiy [21]2 years ago
8 0

The answer is Choice (A)

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For the love of God help me !! I'm desperate for it tomorrow
USPshnik [31]
D:5x-2>0 \wedge x>0 \wedge x-1>0\\D:5x>2 \wedge x>0 \wedge x>1\\D: x>\frac{2}{5} \wedge x>1\\D:x>1\\\log_2(5x-2)-\log_2x-\log_2(x-1)=2\\\log_2\frac{5x-2}{x(x-1)}=\log_24\\\frac{5x-2}{x(x-1)}=4\\4x(x-1)=5x-2\\4x^2-4x=5x-2\\4x^2-9x+2=0\\4x^2-x-8x+2=0\\x(4x-1)-2(4x-1)=0\\(x-2)(4x-1)=0\\x=2 \vee x=\frac{1}{4}\\\frac{1}{4}\not \in D \Rightarrow \boxed{x=2}
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3 years ago
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Y-intercept oftheparabola y = x2 – 2x – 3
inessss [21]
(X-3)(X+1)=0
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6 0
2 years ago
How many ways can a 2 person subcommittee be selected from a comittee of 6 people?
BaLLatris [955]

Answer: 15 ways

Step-by-step explanation: 5+4+3+2 + 1 = 15

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1 year ago
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Solve the equation on the<br> interval [0, 27r).<br> 4(sin x)2 - 2 = 0
Ket [755]

4[sin(x)]^2 - 2 = 0\implies 4[sin(x)]^2=2\implies [sin(x)]^2=\cfrac{2}{4}\implies [sin(x)]^2=\cfrac{1}{2} \\\\\\ sin(x)=\pm\sqrt{\cfrac{1}{2}}\implies sin^{-1}[sin(x)]=sin^{-1}\left( \pm\sqrt{\cfrac{1}{2}} \right)\implies x=sin^{-1}\left( \pm\sqrt{\cfrac{1}{2}} \right)

x=sin^{-1}\left( \pm\cfrac{\sqrt{1}}{\sqrt{2}} \right)\implies x=sin^{-1}\left( \pm\cfrac{1}{\sqrt{2}} \right)\implies x=sin^{-1}\left( \pm\cfrac{\sqrt{2}}{2} \right) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill x=\cfrac{\pi }{4}~~,~~\cfrac{3\pi }{4}~~,~~\cfrac{5\pi }{4}~~,~~\cfrac{7\pi }{4}~\hfill

Check the picture below.

4 0
2 years ago
Evaluate limit:<br> lim x approaches 0 e^(5x)-1-5x/x^2
rosijanka [135]
Lim as x approches 0 of (e^(5x) - 1 - 5x)/x^2 = lim as x approaches 0 of (5e^(5x) - 5)/2x = lim as x approaches 0 of 25e^(5x)/2 = 25/2 = 12.5
8 0
3 years ago
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