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scoundrel [369]

3 years ago

14

A spherical balloon has a 26-in diameter when it is fully inflated. Half of the air is let out of the balloon. Assume that the

balloon remains a sphere.
a. Find the volume of the fully-inflated balloon.
b. Find the volume of the half-inflated balloon.
c. What is the diameter of the half-inflated balloon?
I need help with c

1 answer:

Lady_Fox [76]3 years ago

8 0

Answer:

Volume A = 4/3 x pi x r^3 = 4/3 x 3.14 x 13^3 = 9198.11 cubic inches

Volume B = 4/3 x pi x r^3 = 4/3 x 3.14x 7.5^3 = 1766.25 cubic inches

Diameter = 13 in

Step-by-step explanation:

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“encontrar la integral indefinida y verificar el resultado mediante derivación”

Oliga [24]

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Haz la sustitución:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C$

Para confirmar el resultado:

$\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}$

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Sustituye:

$y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx$

$\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C$

(Te dejaré confirmar por ti mismo.)

$I=\displaystyle\int\frac x{\sqrt{1-x^2}}\,\mathrm dx$

Sustituye:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C$

$I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}$

Sustituye:

$u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}$

$\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C$

Podemos hacer que esto se vea un poco mejor:

$\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}$

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4 0

3 years ago

Which expression can be used to find 7(-6 1/8)

andrezito [222]

The first option,

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7 0

2 years ago

Read 2 more answers

Area of the figure??

8_murik_8 [283]

Answer:

28.5 sq. meters

Step-by-step explanation:

6x3=18 + 6 x 2 divided by 2=6 + 3x3 divided by 2=4.5

=28.5

8 0

3 years ago

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VikaD [51]

Answer:

see explanation

Step-by-step explanation:

Using the cosine and tangent ratios in the right triangle

cos41° = $\frac{adjacent}{hypotenuse}$ = $\frac{VW}{VX}$ = $\frac{7}{VX}$

Multiply both sides by VX

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VX = $\frac{7}{cos41}$ ≈ 9.3

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The sum of the 3 angles in a triangle = 180°

Subtract the sum of the given angles from 180° for ∠ X

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3 0

3 years ago

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Elis [28]

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3 years ago