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BARSIC [14]
3 years ago
15

Which expression is equivalent to (3b +2r) + (4b + r )

Mathematics
2 answers:
VladimirAG [237]3 years ago
6 0

Answer:

Step-by-step explanation:

The one that looks most closely to this.

(3b + 2r) + (4b + r)                 Remove the brackets.

3b + 2r + 4b + r                     Rewrite so the terms are close to each other.

3b + 4b + 2r + r                     Combine

7b + 3r    or

3r + 7b

One of the two of these can be given as an answer. Please give choices if you want me to give an exact answer.

givi [52]3 years ago
5 0

\implies {\blue {\boxed {\boxed {\purple {\sf { \: 7b + 3r}}}}}}

\sf \bf {\boxed {\mathbb {Step-by-step\:explanation:}}}

\: (3b + 2r) + (4b + r)

➼\: 3b + 2r + 4b + r

Combining like terms, we have

➼\: 3b + 4b + 2r + r

➼\: 7b + 3r

\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{❦}}}}}

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Renee is simplifying the expression (7) (StartFraction 13 over 29 EndFraction) (StartFraction 1 over 7 EndFraction). She recogni
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Answer:

The correct option is commutative property.

Step-by-step explanation:

The expression that Renee is simplifying is:

(7)\cdot(\frac{13}{29})\cdot(\frac{1}{7})

It is provided that, Renee recognizes that 7 and \frac{1}{7} are reciprocals, so she would like to find their product before she multiplies by \frac{13}{29}.

The associative property of multiplication states that:

a\times b\times c=(a\times b)\times c=a\times (b\times c)

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a\times b\times c=a\times c\times b=c\times a\times b

The distributive property of multiplication states that:

a\cdot (b+c)=a\cdot b+a\cdot c

The identity property of multiplication states that:

a\times 1=a\\b\times 1=b

So, Renee should use the commutative property of multiplication to find the product of 7 and \frac{1}{7},

(7)\cdot(\frac{13}{29})\cdot(\frac{1}{7})=(7\times\frac{1}{7})\times\frac{13}{29}=\frac{13}{29}

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Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega
AlekseyPX

Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is \frac{1}{3^8} is correct answer

Therefore [(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2

To find the simplified form of the given expression :

[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2

=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2 ( using the property (ab)^m=a^m.b^m )

=(2^6)(3^{-12})\times (2^{-6})(3^4) ( using the property (a^m)^n=a^{mn}

=(2^6)(2^{-6})(3^{-12})(3^4) ( combining the like powers )

=2^{6-6}3^{-12+4} ( using the property a^m.a^n=a^{m+n} )

=2^03^{-8}

=\frac{1}{3^8} ( using the property a^{-m}=\frac{1}{a^m} )

Therefore [(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}

Therefore option "StartFraction 1 Over 3 Superscript 8" is correct

That is \frac{1}{3^8} is correct answer

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3 years ago
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