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Is w=1.47 a solution to the equation #1.23(0.53+w)^2=16#?

Answer 1

Substitute #1.47# for #w# and simplify the left side of the equation. If it is equal to 16 then #w = 1.47# is a solution to the equation. See full process below:

Substitute #color(red)(1.47)# for #color(red)(w)# and simplify the left side of the equation.

#1.23(0.53 + color(red)(w))^2 = 16# substituting and simplifying gives:

#1.23(0.53 + color(red)(1.47))^2 = 16#

#1.23 xx (2)^2 = 16#

#1.23 xx 4 = 16#

#4.92 != 16#

Because 4.92 does not equal 16, #w = 1.47# is not a solution for this equation.

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Answer 2

Isaiah Anthony

To determine if ( w = 1.47 ) is a solution to the equation ( 1.23(0.53 + w)^2 = 16 ), we substitute ( w = 1.47 ) into the equation and then solve for equality.

[ 1.23(0.53 + 1.47)^2 = 16 ] [ 1.23(2)^2 = 16 ] [ 1.23(4) = 16 ] [ 4.92 = 16 ]

( 4.92 \neq 16 )

Thus, ( w = 1.47 ) is not a solution to the equation ( 1.23(0.53 + w)^2 = 16 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.