-2(x+5)=2 is an example of wich property

1 answer:

5 0

-2(x + 5) = 2 |use distributive property: a(b + c) = ab + ac

-2x - 10 = 2 |add 10 to both sides

-2x = 12 |divide both sides by (-2)

x = -6

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A circle centered at (-2,3) and contains point (5,9) which represents the equation

SIZIF [17.4K]

The standard form of the equation of a circle is (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle, (x,y) is a point of the circle, and r is the length of the radius of the circle. When the equation of a circle is written, h,k, and r are numbers, while x and y are still variables. (x-2)^2 + (y-k)^2 = 16 is an example of a circle. The problem gives us two of the three things that a circle has, a point (5,9) and the center (-2,3). We need to find the radius in order to write the equation. We substitute -2 for h, 3 for k, 5 for x, and 9 for y to get (5 - (-2))^2 + (9 - 3)^2 = r^2 We simplify: 49 + 36 = r^2, r^2 = 85. We only need to know r^2 because the equation of a circle has r^2. We now have all the information to write the equation of a circle. (x + 2)^2 + (y - 3)^2 = 85.

8 0

2 years ago

Use spherical coordinates. Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 81, above the xy-plane, an

Natasha_Volkova [10]

Answer:

The volume of the solid is 243 $\sqrt{2} \ \pi$

Step-by-step explanation:

From the information given:

BY applying sphere coordinates:

0 ≤ x² + y² + z² ≤ 81

0 ≤ ρ² ≤ 81

0 ≤ ρ ≤ 9

The intersection that takes place in the sphere and the cone is:

$x^2 +y^2 ( \sqrt{x^2 +y^2 })^2 = 81$

$2(x^2 + y^2) =81$

$x^2 +y^2 = \dfrac{81}{2}$

Thus; the region bounded is: 0 ≤ θ ≤ 2π

This implies that:

$z = \sqrt{x^2+y^2}$

ρcosФ = ρsinФ

tanФ = 1

Ф = π/4

Similarly; in the X-Y plane;

z = 0

ρcosФ = 0

cosФ = 0

Ф = π/2

So here; $\dfrac{\pi}{4} \leq \phi \le \dfrac{\pi}{2}$

Thus, volume: $V = \iiint_E \ d V = \int \limits^{\pi/2}_{\pi/4} \int \limits ^{2\pi}_{0} \int \limits^9_0 \rho ^2 \ sin \phi \ d\rho \ d \theta \ d \phi$