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 one year ago
a table for y=2x^28 is given solve each equation. a. 2x^28=0,b. 2x^28<0,c. 2x^28>0.
 one year ago
a table for y=2x^28 is given solve each equation. a. 2x^28=0,b. 2x^28<0,c. 2x^28>0.

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sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1where is the table?

cdelomas
 one year ago
Best ResponseYou've already chosen the best response.0it is in the question

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.12x^28=0 2x^2=8 x^2=4 x=2 or 2

cdelomas
 one year ago
Best ResponseYou've already chosen the best response.0is that forthe equation a.

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1Did u understand it?

cdelomas
 one year ago
Best ResponseYou've already chosen the best response.0yes for thef irst one I did

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.12x^28<0 2x^2<8 x^2<4 x^2<(+2)^2 Now, IF x^2<(+a)^2 then a<x<a

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1SO, can u complete it?

cdelomas
 one year ago
Best ResponseYou've already chosen the best response.0no do not get that one

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.12x^28<0 2x^2<8 x^2<4 x^2<(+2)^2 IS it OKAY up to here?

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1IF x^2<(+a)^2 then a<x<a

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1x^2<(+2)^2 2<x<2

cdelomas
 one year ago
Best ResponseYou've already chosen the best response.0yes but how do i figure it out

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1HINT: IF x^2<(+a)^2 then a<x<a

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1It is like formula

cdelomas
 one year ago
Best ResponseYou've already chosen the best response.0howdo i figurethef o rmula out

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1U want me to derive it?

cdelomas
 one year ago
Best ResponseYou've already chosen the best response.0hello what doyou m ean

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1I am sorry but I guess u need to revise your algebra once

cdelomas
 one year ago
Best ResponseYou've already chosen the best response.0soyoucannoth elp meany further

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1Sorry not so good at explaining things

cdelomas
 one year ago
Best ResponseYou've already chosen the best response.0ok Iwil l try someelse

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1@mathslover can help u?

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1He is good at explaining

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1I am well at explanation :)

cdelomas
 one year ago
Best ResponseYou've already chosen the best response.0socanyouhelpme mathsolver

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1I will be right back ... wait for 2 minutes please

wio
 one year ago
Best ResponseYou've already chosen the best response.0\[ 2x^28<0 \\ 2x^2<8 \\ x^2<4 \\ \sqrt{x^2}<\sqrt{4} \]Now it is important to remember that \(\sqrt{\quad}\) mean the "positive square root" (not the negative one) and that \(\sqrt{a^2}\) is the definition of of the absolute value.\[ x < 2 \]

wio
 one year ago
Best ResponseYou've already chosen the best response.0To solve absolute value equations, we have to split it up into two cases. Case 1: Assume \(x\) (the thing in the absolute value) is positive. \[ x < 2 \\ x<2 \] That was easy. Case 2: Assume \(x\) is negative: \[ x < 2 \\ x < 2 \\ x > 2 \]Remember that if \(x\) is negative, then \(x =x\) because the absolute value bars had to flip the sign to make \(x\) positive. Also remember that when you multiply/divide both sides of an inequality by a negative number, the equality flips. Hence \(<\) became \(>\) in this instance.

wio
 one year ago
Best ResponseYou've already chosen the best response.0Notation allows us to conveniently combine the expressions \(x < 2\) and \(x > 2\) into \(2 < x < 2\). Does this help, @cdelomas ?

wio
 one year ago
Best ResponseYou've already chosen the best response.0One thing worthy of committing to memory is that: \[ x < a \implies a < x < a \\ x > a \implies x < a,\quad x > a \]

cdelomas
 one year ago
Best ResponseYou've already chosen the best response.0so for equation b is it squareroot x2>squareroot 4.
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