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asambeis [7]
3 years ago
14

What is the answer of (4x+8) (2x-4)

Mathematics
2 answers:
andrey2020 [161]3 years ago
8 0

Answer:

8x² - 32

Step-by-step explanation:

(4x + 8) (2x - 4)

Distribute

(4x)(2x - 4) + (8)(2x - 4)

8x² - 16x + 16x - 32

Cancel out 16x

8x² - 32

Hope this was useful to you!

Zielflug [23.3K]3 years ago
4 0

Answer:

Step-by-step explanation:

you can foil this so

(4x × 2x)+ (4x × -4)+ (8 × 2x)+(8×-4)

when simplified, the answer is

8x^{2} +( -16x )+ 16x +( -32), which when further simplified is

8x^{2} - 32

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Step-by-step explanation:

if A = 1 2 b then it comes to h

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Robert is purchasing some packs of chicken from the supermarket. The first pack weighs 45 ounces. The second pack weighs 3 pound
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Answer: 3 pounds

Step-by-step explanation:

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-3(4u-v)-5(-v+3u)<br> For 20 points
natita [175]

Answer:

−27u + 8v

Explanation:

−3(4u − v) − 5( − v + 3u)

[ Distribute ]

= (−3)(4u) + (−3)(−v) + (−5)(−v) + (−5)(3u)

= −12u + 3v + 5v + −15u

[ Combine Like Terms ]

= −12u + 3v + 5v + −15u

= (−12u + −15u) + (3v + 5v)

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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

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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2 years ago
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