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4 years ago

What is the greatest common factor of 15x^3y^2 and 20x^4y^4?

Mathematics

1 answer:

Vanyuwa [196]4 years ago

6 0

<h3>What is the greatest common factor of 15x³y² and 20x⁴y⁴? </h3>

First we have to find the factors of 15x³y²

15x³y² = 3 * 5x³y²

Then we have to find the factors of 20x⁴y⁴

20x⁴y⁴ = 2²x¹y² * 5x³y²

Now we have to find the common factors to both numbers.

The common factors are 5x³y²

Answer : GCF = (15x³y²,20x⁴y⁴) = 5x³y²

Hope this helps!

$\textit{\textbf{Spymore}}$

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2 years ago

Find equations of the spheres with center(3, −4, 5) that touch the following planes.a. xy-plane b. yz- plane c. xz-plane

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(a) (x - 3)² + (y + 4)² + (z - 5)² = 25

(b) (x - 3)² + (y + 4)² + (z - 5)² = 9

Step-by-step explanation:

The equation of a sphere is given by:

(x - x₀)² + (y - y₀)² + (z - z₀)² = r² ---------------(i)

Where;

(x₀, y₀, z₀) is the center of the sphere

r is the radius of the sphere

Given:

Sphere centered at (3, -4, 5)

=> (x₀, y₀, z₀) = (3, -4, 5)

(a) To get the equation of the sphere when it touches the xy-plane, we do the following:

i. Since the sphere touches the xy-plane, it means the z-component of its centre is 0.

Therefore, we have the sphere now centered at (3, -4, 0).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, -4, 0) as follows;

$d = \sqrt{(3-3)^2+ (-4 - (-4))^2 + (0-5)^2}$

$d = \sqrt{(3-3)^2+ (-4 + 4)^2 + (0-5)^2}$

$d = \sqrt{(0)^2+ (0)^2 + (-5)^2}$

$d = \sqrt{(25)}$

d = 5

This distance is the radius of the sphere at that point. i.e r = 5

Now substitute this value r = 5 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 5²

(x - 3)² + (y + 4)² + (z - 5)² = 25

Therefore, the equation of the sphere when it touches the xy plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 25

(b) To get the equation of the sphere when it touches the yz-plane, we do the following:

i. Since the sphere touches the yz-plane, it means the x-component of its centre is 0.

Therefore, we have the sphere now centered at (0, -4, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (0, -4, 5) as follows;

$d = \sqrt{(0-3)^2+ (-4 - (-4))^2 + (5-5)^2}$

$d = \sqrt{(-3)^2+ (-4 + 4)^2 + (5-5)^2}$

$d = \sqrt{(-3)^2 + (0)^2+ (0)^2}$

$d = \sqrt{(9)}$

d = 3

This distance is the radius of the sphere at that point. i.e r = 3

Now substitute this value r = 3 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 3²

(x - 3)² + (y + 4)² + (z - 5)² = 9

Therefore, the equation of the sphere when it touches the yz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 9

(b) To get the equation of the sphere when it touches the xz-plane, we do the following:

i. Since the sphere touches the xz-plane, it means the y-component of its centre is 0.

Therefore, we have the sphere now centered at (3, 0, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, 0, 5) as follows;

$d = \sqrt{(3-3)^2+ (0 - (-4))^2 + (5-5)^2}$

$d = \sqrt{(3-3)^2+ (0+4)^2 + (5-5)^2}$

$d = \sqrt{(0)^2 + (4)^2+ (0)^2}$

$d = \sqrt{(16)}$

d = 4

This distance is the radius of the sphere at that point. i.e r = 4

Now substitute this value r = 4 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 4²

(x - 3)² + (y + 4)² + (z - 5)² = 16

Therefore, the equation of the sphere when it touches the xz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 16

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3 years ago

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Harlamova29_29 [7]

Answer:

Step-by-step explanation:

AA similarity means 2 angles must be similar.

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