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3 years ago

HELP ME PLEASEEEEEEEEEEEEEEE

Mathematics

2 answers:

Murljashka [212]3 years ago

5 0

Well i'm gonna be honest : my answer probably is wrong, but here ya go :

so since the y intercept of f(x) is -2 translating that 6 units upwards will make it positive 4

so if six units upwards takes it to g(x) the answer should be 4

anygoal [31]3 years ago

4 0

Ok so the formula they gave you is in the form y=mx+b. What's different however is that they say the graph is translated up 6 units. So the formula f(x)=-4x-2 becomes f(x)=-4x+4 (since b represents how high the graph is). And to find any y intercept, you must make X zero, and -4(0) +4 is now equal to 4 and 4 is your answer

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$4\sqrt{3-\frac{1}{x} } -\sqrt{\frac{x}{3x-1} } =3\\4\sqrt{\frac{3x-1}{x} } -\sqrt{\frac{x}{3x-1} } =3\\\\put ~\sqrt{\frac{3x-1}{x}} =a\\4a-\frac{1}{a} =3\\4a^2-1-3a=0\\4a^2-4a+1a-1=0\\4a(a-1)+1(a-1)=0\\(a-1)(4a+1)=0\\a=1,-\frac{1}{4}\\$

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$\frac{(1-2x)^2 \times (x^2-9)}{-x^2-x+6} \geq 0,\\both ~numerator ~and~denominator ~are~of~same~sign.\\\frac{(1-2x)^2(x^2-9)}{-x^2-3x+2x+6} \geq 0\\\frac{(1-2x)^2(x+3)(x-3)}{-x(x+3)+2(x+3)} \geq 0\\\frac{(1-2x)^2(x+3)(x-3)}{(x+3)(-x+2)} \geq 0\\(1-2x)^2\geq 0~(always)\\x\neq -3\\case.~1.\\both~numerator~and~denominator >0\\\frac{x-3}{-x+2} \geq 0\\x-3\geq 0\\x\geq 3\\-x+2> 0\\-x>-2\\x$

case 2.

both numerator and denominator are negative.

$x-3\leq 0\\x\leq 3\\-x+2\leq 0\\-x\leq -2\\x\geq 2\\solution~is~(-\infty,-3)U(-3,3]U[2,\infty)$

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