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Bad White [126]
3 years ago
8

1 and 16/24 + 1 and 21/24

Mathematics
1 answer:
AlexFokin [52]3 years ago
6 0

Answer:

turn into improper fractions

1*24+16=40

40/24

1*24+21=45

45/24

40/24+45/24=85/24

work out how many times 24 goes into 85 to get back to a mixed number.

24 goes into 85 three times with 13 left over so the answer is

3 and 13/24

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icang [17]
Not sure about remainder theorem but I'm sure that the last terms should all multiply tho the last term


see the expanded form is -45

so the last terms of each binomial should multiply to -45

3 times -5 times ?=-45
-15 times ?=-45
divide by -15
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the question mark is 3
5 0
3 years ago
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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

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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 )

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=\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

6 0
3 years ago
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Step-by-step explanation:

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