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3 years ago

13

Spaceship Earth is traveling around the sun at a speed of 18.5 miles per second. How far will we travel during a 52- minute math

class?

2 answers:

Degger [83]3 years ago

8 0

It will travel 57,720 miles.

Amanda [17]3 years ago

3 0

57,720 miles at that speed

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What was the depth of the submarine after 6 minutes

pishuonlain [190]

-120 + 13m, at time m = 6, would be -120 + 78, or -42 feet.

3 0

3 years ago

a 18 ft tall statue standing next to a globe casts a 12 ft shadow. Of the globe casts a shadow that is 2 ft ling, then how tall

Igoryamba

<h3>Answer:</h3>

3 ft

<h3>Step-by-step explanation:</h3>

The statue's height is 1.5 times the length of its shadow, so we expect the same relationship for the globe.

... 1.5 × 2 ft = 3 ft

_____

<em>Comment on the problem</em>

As a practical matter, with the sun high enough in the sky to cast a shadow shorter than the object's height, it will be quite difficult to measure the length of the shadow of the point at the top of the globe. The shadow of other parts of the globe will interfere.

4 0

3 years ago

Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)

Triss [41]

$g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}$

The surface element is

$\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

and the integral is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)$

###

To compute the last integral, you can integrate by parts:

$u=r\implies\mathrm du=\mathrm dr$

$\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}$

$\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr$

For this integral, consider a substitution of

$r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds$

$\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds$

$\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds$

$=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds$

and the result above follows.

4 0

4 years ago

A bag of apples weighs 7 7/8 pounds. By weight, 1/15 of the apples are rotten. What is the weight of the good apples?

siniylev [52]

The good apples are 14/15 of the total apples. So multiply the total apples by the weight of good ones to find the solution. $\frac{63}{8}*\frac{14}{15} = \frac{294}{40} = 7\frac{7}{20}lbs$

4 0

3 years ago

The sum of the first three terms in a GP is 38.Their product is 1728.Find the values of the three terms.

dusya [7]

Answer:

The value of the three terms is 8 and 18

Step-by-step explanation:

Let "a" be the first term and "r" be the common ratio.

Then from the condition, we have these two equations

a + ar + ar^2 = 38, (1)

a*(ar*)*(ar^2) = 1728. (2)

From equation (2), a^3*r^3 = 1728, or (ar)^3 = 1728, which implies

ar = root%283%2C1728%29 = 12; (3)

hence,

r = 12%2Fa. (4)

Now, in equation (1) replace the term ar by 12, based on (3). You will get

a + 12 + ar^2 = 38, which implies

a + ar^2 = 26. (5)

Next, substitute r = 12%2Fa into equation (5), replacing "r" there. You will get

a + a%2A%28144%2Fa%5E2%29 = 26, or

a + 144%2Fa = 26.

Multiply by "a" both sides and simplify

a^2 - 26a + 144 = 0,

%28a-13%29%5E2 - 169 + 144 = 0

%28a-13%29%5E2 = 25

a - 13 = +/- sqrt%2825%29 = +/- 5.

Thus two solutions for "a" are a = 13 + 5 = 18 or a = 13 - 5 = 8.

If a = 8, then from (4) r = 12%2F8 = 3%2F2.

If a = 18, then from (4) r = 12%2F18 = 2%2F3.

In the first case, if a = 8, then the three terms are 8, 8%2A%283%2F2%29 = 12 and 8%2A%283%2F2%29%5E2 = 18.

In this case, the sum of terms is 8 + 12 + 18 = 38, so this solution does work.

In the second case, if a = 18, then the three terms are 18, 18%2A%282%2F3%29 = 12 and 18%2A%282%2F3%29%5E2 = 8.

In this case, the sum of terms is 18 + 12 + 8 = 38, so this solution does work, too.

ANSWER. The problem has two solution:

a) first term is 18; the common difference is 2%2F3 and the progression is 18, 12, 8.

b) first term is 8; the common difference is 3%2F2 and the progression is 8, 12, 18.

4 0

3 years ago