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Sloan [31]
2 years ago
7

How do you solve this (3y+1)^1/2=y-1

Mathematics
1 answer:
Vlad [161]2 years ago
5 0

Answer: y=-3

Step-by-step explanation:

(number)^1 is just (number)

so (3y+1)^1/2=y-1 is (3y+1)/2=y-1

multiply 2 on both side so get rid of fraction

then 3y+1=2(y-1)

3y+1=2y-2

y=-3

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Traces of surface x2 + y2 − z2 = 1. Determine the equation for the family of traces in x = n.
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Answer:

\mathbf{y^2 -z^2 =1- n^2}

Step-by-step explanation:

Give that:

the surface equation is  x^2 +y^2 -z^2 =1

from the family of traces in x = n given that x^2 +y^2 -z^2 =1 , the equation can be represented as :

n^2 +y^2 -z^2 =1

\mathbf{y^2 -z^2 =1- n^2}

This represents a family of hyperbola for all values of n expects that n = ± 1

So, if n = ± 1,

Then

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Step-by-step explanation:

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2 years ago
Rewrite using a single exponent.<br> 53.53
Helga [31]

Answer:

{5}^{6} \:  \: or \:  \:  {25}^{3}  = 15625

Step-by-step explanation:

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Here we can solve this in 2 ways ,

Method 1 :

{5}^{3}  \times  {5}^{3}

Here the bases are equal. So,

=  >   {5}^{3 + 3}  =  {5}^{6}  = 15625

Method 2 :

{5}^{3}  \times {5}^{3}

Here the exponents are same. So,

=  >  {(5 \times 5)}^{3}  =  {25}^{3}  = 15625

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