Question

This is the new question

Answer 1

HELP WITH NUMBER 6 7 8

Answer 2

\[\large 0 = 2[e^{-t}(-1)+(1-t)e^{-t}(-1)]\]divide by 2, take out common factor of e^-t\[\large 0 = e^{-t}[(-1)+(1-t)(-1)]\]set each equal to zero and solve.

Answer 3

no solution

Answer 4

e^-t

Answer 5

The other factor has solutions.

Answer 6

\[(1-t)=0\]
t=1

Answer 7

-1 doesnt =0
no solution
Thats all iget

Answer 8

The other factor was not 1-t... look closer.

7.\[\large 0 = -2x^2(e^{(-x)^2})(-2)-2e^{(-x)^2}\]take out common factor \(-2e^{(-x)^2}\)\[\large 0 = -2(e^{(-x)^2})(x^2(-2)+1)\]

Answer 9

"-1 doesnt =0
no solution
Thats all iget "

none of that makes any sense.

Answer 10

wow i need seroius help with that

Answer 11

\[\large 0 = e^{-t}[(-1)+(1-t)(-1)] \]if one factor is e^-t, then what is the other factor?

Answer 12

1?

Answer 13

im not sure

Answer 14

You can also factor out a -1 \[\large 0 = e^{-t}[(-1)+(1-t)(-1)] \]\[\large 0 = -e^{-t}[1+(1-t)]\]

Answer 15

so from now do u use the zero property rule

Answer 16

Yes. The * separates the factors\[\large 0 = -e^{-t}*[1+(1-t)]\]

Answer 17

so then
0=-e^(-t)

simplify that then

1+(1-t)=0
-1 -1
(1-t)=-1
-t=-2
t=2

Answer 18

right?

Answer 19

Yes

Answer 20

how would i then simplify then 0=e....

Answer 21

i think theres no solution bc theres a 0

Answer 22

There is a solution. You just found it.

Answer 23

Back to 7.\[\large 0 = -2(e^{(-x)^2})[x^2(-2)+1]\]solve.

Answer 24

i got one answer which is 2. bc i did the zero property.

Answer 25

THIS IS WHAT I MEAN

Answer 26

u see the question mark thats what i was talking about. will that be no solution

Answer 27

n for number 7
\[0=x^2 (e^(-x)^2)(-1)-e^(-x)^2\]

Answer 28

e^t never equals zero. Onto number 7.

Answer 29

ah i got u nowwww :)

Answer 30

so i divided by -2

Answer 31

Already factored 7, see earlier post \[\large 0 = -2(e^{(-x)^2})[x^2(-2)+1]\]

Answer 32

so then
\[-2x^2+1=0\]

Answer 33

I assume you can solve that. Onto 8.\[\large 0 = -2 w^2 e^{(-w)^2}+e^{(-w)^2}\]factor out \(e^{(-w)^2}\)\[\large 0 = e^{(-w)^2}(-2 w^2+1)\]

Answer 34

\[-2x^2=-1\]

Answer 35

thats for number 7. ityped that 10 mins ago n it just sent. FREAKING WIFI

Answer 36

\[x=\sqrt{1/2}\]

Answer 37

for number 7^

Answer 38

You forgot the plus or minus sign, when taking a square root.

Answer 39

8.
\[w ^{2}= 1/2\]

Answer 40

oh yeah 7. x= plus or minus \[\sqrt{1/2}\]

Answer 41

8. w= plus or minus radiccal 1/2

Answer 42

what u think?

Answer 43

THANK U SO MUCH FOR THE HELP @agent0smith

Answer 44

do u mind helping me with some other like word problems.??

Answer 45

its about average rate of change

Answer 46

Rationalize it\[\large \pm \sqrt{\frac{ 1 }{ 2 }} = \pm \frac{ 1 }{ \sqrt 2} = \pm \frac{ \sqrt 2 }{ 2 }\]I'm hungry though so maybe.