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 2 years ago
a table for y=2x^28 is given solve each equation. a. 2x^28=0,b. 2x^28<0,c. 2x^28>0.
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1where is the table?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

cdelomas
 2 years ago
Best ResponseYou've already chosen the best response.0socanyouhelpme mathsolver

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

wio
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0so for equation b is it squareroot x2>squareroot 4.
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