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4 years ago

The GCF of 28x^3 and 16x^2y^2

Mathematics

2 answers:

AlexFokin [52]4 years ago

7 0

Answer:

$4x^2$

Step-by-step explanation:

We are asked to find the GCF of $28x^3$ and $16x^2y^2$ .

The greatest common factor of numerical part would be GCF of 28 and 16.

Factors of 28: 1, 2, 4, 7, 14, 28.

Factors of 16: 1, 2, 4, 8, 16.

The greatest common factor of 28 and 16 is 4.

The GCF of variable part:

$x^3=x\cdot x\cdot x$

$x^2y^2=x\cdot x\cdot y\cdot y$

The greatest common factor of $x^3$ and $x^2y^2$ is $x^2$ .

Upon combining the GCF of numerical part and algebraic part, we will get: $4x^2$

Therefore, the greatest common factor of both expressions is $4x^2$ .

masya89 [10]4 years ago

6 0

I think the greatest common factor is 4

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Mars2501 [29]

Answer:

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Step-by-step explanation:

speed=distance/time

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3 years ago

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ASHA 777 [7]

Answer:

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Step-by-step explanation:

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3 years ago

Shane is using pieces of square paper to cover a cube shaped gift with sides that are 9 inches long to find out the area of the

nordsb [41]

Answer:

The area of the cube is 486 inches^2

Step-by-step explanation:

In this question, we are tasked with calculating the area of cube with side 9 inch.

Mathematically, the area can be calculated using the formula A = 6s^2

Now, what we need to do is to substitute 9 inches for s

Thus, A = 6 * 9^2

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3 0

3 years ago

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

6 0

3 years ago

Read 2 more answers

The square (5x² + 6xy)² is

kari74 [83]

Answer:

${25x}^{4} + 60 {x}^{3} y + 36 {x}^{2} {y}^{2}$

Step-by-step explanation:

$(5 {x}^{2} )^{2} + 2 \times 5 {x}^{2} \times 6xy + (6xy)^{2}$

Gives the above answer

8 0

3 years ago

Read 2 more answers