Powers and Roots: Identify the greater value

Indicates the corresponding sign:

^{$-3+(\sqrt{100}-3^2-1^{14}):30+3\text{ }\text{\textcolor{red}{\_\_}}\text{ }6^2:6\cdot(3-2)-6$}

According to the given problem, whether it's an addition or subtraction each of the digits that come up separately,

this is done within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are done before division and subtraction, and that the preceding operations are done before all,

A. We start with the digits that are left in the given problem:

$-3+(\sqrt{100}-3^2-1^{14}):30+3$First, we simplify the digits that are in the denominators on which the division operation takes place, this is done in accordance with the order of operations mentioned, keeping in mind that multiplication precedes subtraction, therefore, we first calculate their numerical values of the numerators in the multiplication (this within the definition of the root as a power, the root of which is a power for everything), subsequently we perform the subtraction operation which is within the denominators and finally we perform the division operation that takes place on the denominators:

$-3+(\sqrt{100}-3^2-1^{14}):30+3 =\\ -3+(10-9-1):30+3 =\\ -3+0:30+3 =\\ -3+0+3 \\$In the last step we mentioned that dividing the number 0 by any number (different from zero) will yield the result 0, we continue with the simple digits we received in the last step and perform the addition operation:

$-3+0+3 =\\ 0$We finish the simple digits that are left in the given problem, we summarize the steps of the simplification:

We received that:

$-3+(\sqrt{100}-3^2-1^{14}):30+3 =\\ -3+0:30+3 =\\ 0$

B. We continue and simplify the digits that are right in the given problem:

$6^2:6\cdot(3-2)-6$In this part, in the first step we simplify the digits within the framework of the order of operations,

In this digit, multiplication takes place on digits in the denominators, therefore, we first simplify this digit, in the process we calculate the numerical values of the numerator in the multiplication which is the divisor in the first digit from the left in the given digit:

$6^2:6\cdot(3-2)-6 =\\ 36:6\cdot1-6 \\$We continue and remember that multiplication and division precede addition and subtraction, keeping in mind that there is no predefined precedence between multiplication and division operations originating from the order of operations mentioned, therefore, we calculate the numerical values of the numerator in the first digit from the left (including the multiplication and division operations) within the execution of one operation after another according to the order from left to right (this is the natural order of operations), subsequently, we complete the calculation within the execution of the subtraction operation:

$36:6\cdot1-6 =\\ 6\cdot1-6 =\\ 6-6 =\\ 0$We finish the simple digits that are right in the given problem, we summarize the steps of the simplification:

We received that:

$6^2:6\cdot(3-2)-6 =\\ 36:6\cdot1-6 =\\ 0$We return to the original problem, and we present the results of the simplifications reported in A and B:

$-3+(\sqrt{100}-3^2-1^{14}):30+3\text{ }\text{\textcolor{red}{\_\_}}\text{ }6^2:6\cdot(3-2)-6 \\ \downarrow\\ 0\text{ }\text{\textcolor{red}{\_\_}}\text{ }0$As a result, we have that:

$0 \text{ }\textcolor{red}{=}0$Therefore, the correct answer here is answer B.

$=$

Indicates the corresponding sign:

^{$\frac{1}{5}\cdot((5+3:3-8)^2:\sqrt{4}-2)\text{ }_{\textcolor{red}{——}\text{\textcolor{red}{ }}}8-(3^2+1)\cdot\frac{1}{10}$}

According to the given problem, whether it is discussed in addition or subtraction each of the terms that comes in its turn,

this is done within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are performed before all others,

A. We start with the terms that are on the left in the given problem:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)$First, we simplify the terms that are in the denominators (the divisors) on which the multiplication operation is performed, this in accordance with the order of operations mentioned, we note that this term will change in terms that are in the denominators (the numerators) on which division is performed, therefore we start with simply the terms that are in these denominators, remembering that division precedes multiplication and subtraction, therefore the beginning will perform the division operation that is in this term and then perform the operations of multiplication and subtraction:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)\\ \frac{1}{5}\cdot\big((5+1-8)^2:\sqrt{4}-2\big)=\\ \frac{1}{5}\cdot\big((-2)^2:\sqrt{4}-2\big)\\$Since the results of the operations of multiplication and subtraction that are in the numerators the level will come out smoothly on these denominators, we continue and perform the strength on the term that is in these denominators, this within that we remember thatthe raising of any number (positive or negative) in a double strength will give a positive result, in contrast we will consider its numerical value of the other side that in strength he is the divisor that is in the term that within the denominators the divisors that were left (this within that we remember that in defining the root as strength, the root he is strength for everything):

$\frac{1}{5}\cdot\big((-2)^2:\sqrt{4}-2\big)=\\ \frac{1}{5}\cdot\big(4:2-2\big)\\$We continue and finish simply the term, we remember that division precedes subtraction and therefore the beginning will calculate the result of the division operation that in the term and then perform the operation of subtraction:

$\frac{1}{5}\cdot\big(2-2\big)=\\ \frac{1}{5}\cdot0=\\ 0$In the last stage we performed the doubling that was left (it is the doubling that is performed on the term that in the denominators), this within that we remember thatthe result of doubling any number (different from zero) in zero here zero,

$$We finished simply the term that is on the left in the given problem, we will summarize the stages of simply:

We received that:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)\\ \frac{1}{5}\cdot\big((-2)^2:\sqrt{4}-2\big)\\ \frac{1}{5}\cdot\big(2-2\big)=\\ 0$

B. We continue and simplify the term that is on the right in the given problem:

$8-(3^2+1)\cdot\frac{1}{10}$In this part to do in the first part simplify the term within the framework of the order of operations,

In this term a doubling that is performed on the term that in the denominators, therefore we simplify first this term, this is done in accordance with the order of operations mentioned, therefore we start from considering its numerical value that in strength that in this term and then perform the operation of multiplication:

$8-(3^2+1)\cdot\frac{1}{10} =\\ 8-(9+1)\cdot\frac{1}{10}=\\ 8-10\cdot\frac{1}{10}\\$We continue and simplify the term that was received in the first stage, we remember in this that multiplication and division precede addition and subtraction, therefore the beginning will perform the doubling in the break, this within that we remember that the doubling in the break means the doubling in the amount of the break, then perform the operation of division that of the break, this is done by appointment, in the last stage will perform the operation of subtraction that remained:

$8-10\cdot\frac{1}{10}=\\ 8-\frac{10\cdot1}{10}=\\ 8-\frac{\not{10}}{\not{10}}=\\ 8-1=\\ 7$We finished simply the term that is on the right in the given problem, we will summarize the stages of simply:

We received that:

$8-(3^2+1)\cdot\frac{1}{10} =\\ 8-10\cdot\frac{1}{10}\\ 7$We return now to the original problem, and we will present the results of simply the terms that were reported in A and in B:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)\text{ }_{\textcolor{red}{——}\text{\textcolor{red}{ }}}8-(3^2+1)\cdot\frac{1}{10} \\ \downarrow\\ 0\text{ }\text{\textcolor{red}{\_\_}}\text{ }7$Therefore the correct answer here is answer A.

$\ne$

Indicates the corresponding sign:

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2\text{ }\textcolor{red}{_—}(5^2-3+6):7\cdot\frac{1}{4}$

For a given problem, whether it involves addition or subtraction each of the terms that come into play separately,

this is done within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are performed before division and subtraction, and that the preceding operations are performed before all others,

A. We will start with the terms that are on the left in the given problem:

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2$First, we simplify the terms that are in the denominators in accordance with the order of operations, this is done by calculating the numerical value of the denominator in strength (this within that we remember that in defining the root as strong, the root itself is strong for everything) and then perform the operation of division and subtraction:

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2 =\\ \frac{1}{\sqrt{16}}\cdot(125+3-4):2^2 =\\ \frac{1}{\sqrt{16}}\cdot124:2^2$Next, we calculate the numerical values of the part that was passed in strength (practically, if we were to represent the operation of division as broken, this part would have been in the broken position) and as such the numerical value of the part in strength that in the broken position in the terms, next we perform the operations of multiplication and division:

$\frac{1}{\sqrt{16}}\cdot124:2^2 =\\ \frac{1}{4}\cdot124:4 =\\ \frac{1\cdot124}{4}:4=\\ \frac{\not{124}}{\not{4}}:4=\\ 31:4=\\ \frac{31}{4}=\\ 7\frac{3}{4}$In the final stages, we performed the multiplication of the number 124 in break, this we did within that we remember that multiplication in break means multiplication in the broken position, next we performed the operation of division of the break (by condensing the break) and in the final stage we performed the operation of division in the number 4, this operation resulted in a complete answer, and therefore we marked it as break (a break from understanding, an assumption that the position is greater than the position) and in the continuation we converted the break from understanding to a mixed break, by extracting the wholes (the answer to the question: "How many times does the division enter the divisor?") and adding the remaining division to the divisor,

We finished simplifying the terms that are on the left in the given problem, we will summarize the simplification stages:

We received that:

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2 =\\ \frac{1}{\sqrt{16}}\cdot124:2^2 =\\ \frac{1\cdot124}{4}:4=\\ 7\frac{3}{4}$

B. We will continue and simplify the terms that are on the right in the given problem:

$(5^2-3+6):7\cdot\frac{1}{4}$In this part, to simplify the terms within the framework of the order of operations,

In this term, the operation of division of the beginning on terms that are in the denominators, therefore we will first simplify this term,

Let's note that multiplication precedes addition and subtraction, which precede division and subtraction, therefore we will start by calculating the numerical value of the part in strength that in this term, next we perform the operations of division and subtraction:

$(5^2-3+6):7\cdot\frac{1}{4} =\\ (25-3+6):7\cdot\frac{1}{4} =\\ 28:7\cdot\frac{1}{4}$We will continue and simplify the received term, noting that between multiplication and division there is no precedence defined in the order of operations, we perform the operations in this term one after the other according to the order from left to right, which is the natural order of operations:

$28:7\cdot\frac{1}{4} =\\ 4\cdot\frac{1}{4} =\\ \frac{4\cdot1}{4}=\\ \frac{\not{4}}{\not{4}}=\\ 1$ In the second stage, we performed the multiplication in break, this within that we remember (again) that multiplication in break means multiplication in the broken position, in the next stage we performed the operation of division of the break (by condensing the break).

We finished simplifying the terms that are on the right in the given problem, we will summarize the simplification stages:

We received that:

$(5^2-3+6):7\cdot\frac{1}{4} =\\ 28:7\cdot\frac{1}{4} =\\ \frac{\not{4}}{\not{4}}=\\ 1$We return to the original problem, and we will present the results of simplifying the terms that were reported in A and B:

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2\text{ }\textcolor{red}{_—}(5^2-3+6):7\cdot\frac{1}{4} \\ \downarrow\\ 7\frac{3}{4} \text{ }\textcolor{red}{_—}1$As a result, we receive that:

$7\frac{3}{4} \text{ }\textcolor{red}{\neq}1$Therefore, the correct answer here is answer B.

$\ne$