Simplify the following expression: #101-{[(110 -: 2)-: 11]xx(10+4xx2) +7}+ [8xx(20 -: 5-1)-3xx3] -: 5#?

Answer 1

Take your time and methodically go through each bracket and you'll eventually get to #7#

Wow... that's one big equation. Let's take this step by step.

First we'll start with the original:

#101-{[(110-:2)-:11]xx(10+4xx2)+7}+[8xx(20-:5-1)-3xx3]-:5#
Before we dive into this thing, let's look at the structure - there is #101# - big brackets + smaller brackets#-:5#. PEDMAS has us work things in brackets (Parentheses) first and since the big brackets and the smaller brackets are separated by the #+#, we can work them separately. I'm going to simplify the big brackets first:
#{[(110-:2)-:11]xx(10+4xx2)+7}#

There are brackets (and brackets within brackets) in this, so I'm going to work those first. There are 2 sets here and I'll work them side by side. In this first step, let's do the division and in the second set we have both addition and multiplication - so we'll do the multiplication first:

#{[55-:11]xx(10+8)+7}#

I can now do the next division in the first bracket and do the addition in the second:

#{5xx18+7}#

We'll finish this up with the multiplication first, then the addition:

#90+7=97# which I'll substitute back into our original:
#101-97+[8xx(20-:5-1)-3xx3]-:5#

Let's now work that second bracket:

#[8xx(20-:5-1)-3xx3]#

There's a bracket in here that we need to work first. Within that bracket there is both division and subtraction - we'll divide first:

#[8xx(4-1)-3xx3]#

and now the subtraction:

#[8xx3-3xx3]#

We now have 2 multiplications and a subtraction, so we'll do the multiplications first:

#[24-9]=15#

Let's substitute that into the original:

#101-97+15-:5#

Almost there! We have subtraction, addition, and division. We'll do the division first:

#101-97+3#

and now the subtraction and addition:

#4+3=7#
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Answer 2

=#color(magenta)(101)color(green)(+3)color(blue)(-87)#

=#7#

Count the number of terms and work through each one carefully. Each term must give a number answer.

There are only 3 terms in this expression:

#color(magenta)(101)color(blue)(-{[(110-:2)-:11]xx(10+4xx2)+7})color(green)(+[8xx(20-:5-1)-3xx3]-:5)#
Let's handle one at a time. The first is easy. it is #color(magenta)(101)#.
#color(blue)(-{[color(red)((110-:2))-:11]xx(10+color(red)(4xx2))+7})#
#"parentheses first, but remember to do the multiplication "# #"and division before addition and subtraction"#
#color(blue)(-{[color(red)(55)-:11]xx(10+color(red)(8))+7})#
#color(blue)(-{[color(red)(5]xx(color(red)(18))+7})#
#color(blue)(-[color(red)(90+7)] = -97#

Now for the third term:

#color(green)(+[8xx(20-:5-1)-3xx3]-:5)#
#color(green)(+[8xxcolor(red)((20-:5-1))-color(orange)(3xx3)]-:5)#
#color(green)(+[8xxcolor(red)((4-1))-color(orange)(9)]-:5)#
#color(green)(+[8xxcolor(red)(3)-color(orange)(9)]-:5)#
#+[color(red)(24)-color(orange)(9)]-:5)#
#+color(green)(15-:5) = color(green)3#
the whole expression simplifies to #color(magenta)(101)color(blue)(-97)color(green)(+3)#
=#color(magenta)(101)color(green)(+3)color(blue)(-97)#
=#7#
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Answer 3

To simplify the given expression:

  1. ( \frac{110}{2} = 55 )
  2. ( 55 ÷ 11 = 5 )
  3. ( 10 + (4 \times 2) = 10 + 8 = 18 )
  4. ( 5 \times 2 = 10 )
  5. ( 20 ÷ (5 - 1) = 20 ÷ 4 = 5 )
  6. ( 3 \times 3 = 9 )

Now substitute these values back into the expression and simplify step by step:

(101 - \left[ \left( 55 - 5 \right) \times \left( 18 + 7 \right) + 7 \right] + \left[ 8 \times \left( 5 - 3 \times 3 \right) ÷ 5 \right])

(101 - \left[ 50 \times 25 + 7 \right] + \left[ 8 \times (5 - 9) ÷ 5 \right])

(101 - \left[ 1250 + 7 \right] + \left[ 8 \times (-4) ÷ 5 \right])

(101 - 1257 + \left[ -32 ÷ 5 \right])

(101 - 1257 + (-6.4))

(-1156.4)

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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.

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.

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.

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.

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