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Mathematics
b1az3:

(5^(p^(2))-: (5^(p))^(2)) (1)/(p)=125 For the following equation, find the value of . In your final answer, include all of the steps and calculations necessary to solve for.

Mudasir:

First is barket close.

Mudasir:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Mudasir First is barket close. \(\color{#0cbb34}{\text{End of Quote}}\) First step is barket close

b1az3:

ok

Mudasir:

Please attach the file of this question so I understand

b1az3:

ok hold up

b1az3:

1 attachment
Mudasir:

5p^2 is same

Mudasir:

5p^2÷ 1/p(25p)^2 =5p^2×p/625p^2 =5p^2×1/625p =P/125 125 Multiply both side. P/125×125=125 P=125 Simply of your question Check all steps

b1az3:

thanks i suck at this

Florisalreadytaken:

Im kinda confused On what the equation should be — is the one attached right?

Florisalreadytaken:

@Mudasir, according to my question, as long as you solved it, you ought to have a yes or no right?

Mudasir:

Can I solve it.\(\color{#0cbb34}{\text{Originally Posted by}}\) @Florisalreadytaken @Mudasir, according to my question, as long as you solved it, you ought to have a yes or no right? \(\color{#0cbb34}{\text{End of Quote}}\) Can I solve it

Florisalreadytaken:

Certainly

Mudasir:

X=5 not 125

Florisalreadytaken:

for a start, that's a linear equation, which means that there are going to be 2 solutions.

Florisalreadytaken:

Then we can map these two points and draw a line through them.

Florisalreadytaken:

so the solution is going to be a linear line, not just p=125 nor p=5 lol.

Mudasir:

Yes your right sir

Florisalreadytaken:

|dw:1615281613903:dw|

b1az3:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Florisalreadytaken Created with RaphaëlReply Using Drawing \(\color{#0cbb34}{\text{End of Quote}}\) thanks

Florisalreadytaken:

|dw:1615282295042:dw|

Florisalreadytaken:

i used X instead of p, as i was confusing myself using p. You just replace the xs with p.

Florisalreadytaken:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @b1az3 \(\color{#0cbb34}{\text{End of Quote}}\) Here it doesn't say you've got to do the graphing for that equation, so well, I guess we're finished lol.

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