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Can someone please help its like really important and due today please!

Can someone please help its like really important and due today please!-example-1
Can someone please help its like really important and due today please!-example-1
Can someone please help its like really important and due today please!-example-2
User Ted Warner
by
4.0k points

1 Answer

5 votes

25.


b = (2)/(10) = (1)/(5) \\

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26 .


c = (5)/(6) \\

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27 .


d = (6)/(6) = 1 \\

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31 .


x - 3 = \sqrt{ {x}^(2) - 6x + 9}


x - 3 = \sqrt{ ({x - 3})^(2) }


x - 3 = |x - 3|

Thus :


x - 3 = x - 3


x = infinite \: solutions

OR


x - 3 = - (x - 3)


x - 3 = - x + 3


2x = 6


x = 3

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32.


10 = 2 √(3 - 2t)

Divide both sides by 2


(10)/(2) = (2 √(3 - 2t) )/(2) \\


5 = √(3 - 2t)


({5})^(2) = ({ √(3 - 2t) })^(2)


25 = |3 - 2t|

Switch sides


|3 - 2t| = 25

Thus :


3 - 2t = 25

Subtract both sides 3


3 - 3 - 2t = 25 - 3


- 2t = 22

Divide both sides by -2


( - 2t)/( - 2) = (22)/( - 2) \\


t = - 11 \: \: \: acceptable

Or


3 - 2t = - 25

Subtract both sides 3


3 - 3 - 2t = - 25 - 3


- 2t = - 28

Divide both sides by -2


( - 2t)/( - 2) = ( - 28)/( - 2) \\


t = 14 \: \: \: unacceptable

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33.


3 - ({2p - 4})^{ (1)/(3) } = 2

Subtract both sides 3


3 - 3 - \sqrt[3]{2p - 4} = 2 - 3


- \sqrt[3]{2p - 4} = - 1


\sqrt[3]{2p - 4} = 1


({ \sqrt[3]{2p - 4} })^(3) = ({1})^(3)


2p - 4 = 1

Add both sides 4


2p - 4 + 4 = 1 + 4


2p = 5

Divide both sides by 2


(2p)/(2) = (5)/(2) \\


p = (5)/(2) \\

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34.


r = √( - 1 - 2r)


({r})^(2) = ({ √( - 1 - 2r) })^(2)


{r}^(2) = | - 1 - 2r|


{r}^(2) = - 1 - 2r


{r}^(2) + 2r + 1 = 0


({r + 1})^(2) = 0


r + 1 = 0


r = - 1

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35.


√(m + 2) = √(2m)


({ √(m + 2) })^(2) = ({ √(2m) })^(2)


|m + 2| = |2m|

Thus :


m + 2 = 2m

Switch sides


2m = m + 2

Subtract both sides m


2m - m = m - m + 2


m = 2 \: \: acceptable

Or


2m = - (m + 2)


2m = - m - 2

Add both sides m


2m + m = - m + m - 2


3m = - 2

Divide both sides by 3


(3m)/(3) = ( - 2)/(3) \\


m = - (2)/(3) \: \: unacceptable \\

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36 .


\sqrt[3]{4x + 1} - \sqrt[3]{6x - 9} = 0


\sqrt[3]{4x + 1} = \sqrt[3]{6x - 9}


({ \sqrt[3]{4x + 1} })^(3) = ({ \sqrt[3]{6x - 9} })^(3)


4x + 1 = 6x - 9

Switch sides


6x - 9 = 4x + 1

Subtract both sides 4x


6x - 4x - 9 = 4x - 4x + 1


2x - 9 = 1

Add both sides 9


2x - 9 + 9 = 1 + 9


2x = 10

Divide both sides by 2


(2x)/(2) = (10)/(2) \\


x = 5

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The last one is yours bro you got this ....

Have a great time ♡♡

User Jack Dempsey
by
4.0k points