Let $x=4$ and $y=-2$ . Evaluate $(x^2-y^2(10-y^2)-:3)^2$ . Apparently I have to put a question mark here ?

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Subject: Prealgebra
Focus: Squares

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Answer 1 · 2018-06-10T23:50:01

It reduces to 64

For questions of this type, we take the given values $(x=4, y=-2)$ and substitute them into the expression to see what it simplifies to:

$(x^2-y^2(10-y^2)-:3)^2$

$(4^2-(-2)^2(10-(-2)^2)-:3)^2$

Now that the values are placed, we now have to work through the order of operations:

$color(red)(P)$ - Parentheses (also known as Brackets)
$color(blue)(E)$ - Exponents
$color(green)(M)$ - Multiplication
$color(green)(D)$ - Division (this has the same weight as M and so I gave it the same colour)
$color(brown)(A)$ - Addition
$color(brown)(S)$ - Subtraction - (again, same weight as A and so the same colour)

The bracket containing the $(10-(-2)^2)$ term should be done first:

$(10-(-2)^2)$

We first square the $-2$ , then subtract that result from 10:

$(10-4)=6$

$:. (4^2-(-2)^2(6)-:3)^2$

Let's now do the two squares remaining within the bracket:

$(16-4(6)-:3)^2$

Next we have the multiplication and division:

$(16-24-:3)^2$

$(16-8)^2$

We can now do the subtraction and then the square:

$8^2=64$