Axis of symmetry help needed! Algebra

Following decimals into words 15.06

melamori03 [73]

Answer:

Fifteen and six hundredths

Step-by-step explanation:

6 0

3 years ago

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Please help me find X.

Anna007 [38]

It's about 54.3 (repeating)

6 0

3 years ago

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Move the decimal point in 250,000 to The left as many places as necessary to. Find a number that is greater than or equal to 1 a

kotykmax [81]

Your answer is 2.50 (And the rest of the 0s, but they aren't needed...) Since that is greater then 1, but less than 10.

8 0

3 years ago

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

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

$V = \bigg [-cos \phi \bigg]^{\pi/2}_{\pi/4} \bigg [\theta \bigg]^{2 \pi}_{0} \bigg [\dfrac{\rho^3}{3} \bigg ]^{9}_{0}$

$V = [ -0+ \dfrac{1}{\sqrt{2}}][2 \pi -0] [\dfrac{9^3}{3}- 0 ]$

V = 243 $\sqrt{2} \ \pi$

4 0

2 years ago

Nancy found that x=1 is one solution to the quadratic equation (x 2)2=a. what is the value of a? -9 -3 3 9

AlladinOne [14]

The value of the <em>a, </em>in the provided quadratic equation for which Nancy found one solution as x=1 is 9.

<h3>What is the solution of equation?</h3>

The solution of the quadratic equation is the solution of the variable of the equation, for which the equation satisfies.

Nancy found that x=1 is one solution to the quadratic equation. The quadratic equation is,

$(x +2)^2=a$

For this equation, one solution is <em>x</em>=1. Put this value in the above equation to get the value of <em>a,</em>

$(1 +2)^2=a\\(3)^2=a\\9=a\\a=9$

Thus, the value of the <em>a, </em>in the provided quadratic equation for which Nancy found one solution as x=1 is 9.

Learn more about the solution of the equation here;

brainly.com/question/21283540

5 0

1 year ago

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