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3 years ago

9

Which statements about the values of 0.034 and 3.40 are true. Circle ALL that apply.

1 answer:

solong [7]3 years ago

8 0

<span>E.340 is 1/10 0f 0.0340
Hope this helped</span>

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The sum of a number and its additive inverse is equal to?

Usimov [2.4K]

For example, the additive inverse of 12 is –12. The additive inverse of –3 is 3. Formally, the additive inverse of x is –x. Note: The sum of a number and its additive inverse is 0.

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3 years ago

1. Jan spends part of her year as a member of a

antiseptic1488 [7]

Answer:

13 months in a half......

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3 years ago

Last week salazar played 13 more tennis games than perry. if they played a combined total of 53 games. How many games did salaza

d1i1m1o1n [39]

If last week Salazar played 13 more tennis games than Perry and they played a combined total of 53 games, then Salazar played a total of 33 games.

Let the total number of games played by Perry be x.

It is given that, Salazar played 13 more tennis games than Perry.

⇒ Total games played by Salazar = x + 13

Also, Salazar and Perry played a combined of 53 games.

Hence, total number of tennis games played by Salazar and Perry = 53

⇒ Games played by Salazar + Games played by Perry = 53

⇒ x + (x + 13) = 53

2x + 13 = 53

2x = 53 - 13

2x = 40

x = 40 / 2

x = 20

Therefore, total number of games played by Salazar = x+13

= 20 + 13

= 33

Thus, Salazar played total 33 games.

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2 years ago

HELP Please! Can someone solve any of these problems?

enyata [817]

Answer:

i love brent faiyaz

Step-by-step explanation:

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2 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

3 years ago