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3 years ago

Find the value of x

Mathematics

1 answer:

antoniya [11.8K]3 years ago

3 0

Answer:

C. X = 10

Step-by-step explanation:

2x - 17 + (-8) + 2x = 15

4x + (-25) = 15

4x = 40

x = 10

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What's 30-3/5???????

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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

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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:

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

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

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This implies that:

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

ρcosФ = ρsinФ

tanФ = 1

Ф = π/4

Similarly; in the X-Y plane;

z = 0

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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 $\sqrt{2} \ \pi$

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2 years ago

In a missile-testing program, one random variable of interest is the distance between the point at which the missile lands and t

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Answer:

Step-by-step explanation:

From the given data

we observed that the missile testing program

Y1 and Y2 are variable, they are also independent

We are aware that

$(Y_1)^2 and (Y_2)^2$ have $x^2$ distribution with 1 degree of freedom

and $V=(Y_1^2)+(Y_2)^2$ has x^2 with 2 degree of freedom

$F_v(v)=\frac{e^{-\frac{v}{2}}}2$

Since we have to find the density formula

$U=\sqrt{V}$

We use method of transformation

$h(V)=\sqrt{U}\\\\=U$

There inverse function is $h^-^1(U)=U^2$

We derivate the fuction above with respect to u

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Therefore,

$F_v(u)=F_v(h-^1)(u)\frac{dh^-^1}{du} \\\\=\frac{e^-\frac{u^-^}{2} }{2} (2u)\\\\=e^-{\frac{u^2}{2} }U$

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3 years ago

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Answer:

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Step-by-step explanation:

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