All categories

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

You might be interested in

Calculate the cost for using 9,5 in one month and then add vat 15%to the bill

Molodets [167]

its your answer see and mark me brainliest plz

5 0

3 years ago

Find the slope of the curve below at the given points. Sketch the curve along with its tangents at these points. R = sin 2 theta

mr Goodwill [35]

Answer:

-1 

Step-by-step explanation:

Given that:

r = sin 2θ , θ = ± π/4 , ± 3π/4

Recall that:

x = r cosθ

y = r sinθ

The differential of y with respect to x

$\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d \theta}}{\dfrac{dx}{d \theta}}$

$\dfrac{dy}{dx} =\dfrac{\dfrac{dy}{d \theta}}{\dfrac{dx}{d \theta}} = \dfrac{ r cos \theta + sin \theta* \dfrac{dr}{d \theta} } {\dfrac{dr}{d \theta} *cos \theta- sin \theta \ r}$

at θ = π/4 , r = sin π/2

r = 1

$\dfrac{dy}{dx} = \dfrac{ r cos \theta + 2 cos (2 \theta)*sin \ \theta } {2 \ cos (2 \theta) *cos \theta- sin 2 \theta * sin \theta}$

where;

θ = π/4

$\dfrac{dy}{dx} = \dfrac{ 1 \times cos (\dfrac{\pi}{4}) + 2 cos (\dfrac{\pi}{2})*sin (\dfrac{\pi}{4}) } {2 \ cos (\dfrac{\pi}{2})*cos (\dfrac{\pi}{4})- sin (\dfrac{\pi}{2}) * sin (\dfrac{\pi}{4})}$