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4^0-5(-2)^3 I 4-6 I - SQUARE ROOT OF 64

Mathematics
Tutor: None Selected Time limit: 1 Day

Please help me with this expression. thanks

Oct 22nd, 2014

4^0 - 5*(-2)^3 * |4-6| - sqrt(64)

Ok.  lets break this into 4 sections.  4^0, 5(-2)^3, abs(4-6), and sqrt(64)

Section 1.

4^0

As a rule, any number raised to a 0 power is 1 so

4^0 = 1

Section 2

5(-2)^3

Ok, so the first thing we need to do is raise -2 to the 3rd power

-2*-2*-2 = -8

so now we have 5 * -8

5*-8= -40

Section 3

|4-6|

4-6=-2

|-2| = 2

Section 4

sqrt(64) = 8

Now we plug these values back into the equation so

1 - (-40) * 2 - 8 = ?

using order of operations, we get 1 - (-80) - 8 = ?

1 + 80 - 8 =?

= 73

Oct 22nd, 2014

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