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1 year ago

PLS HELP!What is the difference of

Mathematics

1 answer:

serious [3.7K]1 year ago

8 0

Answer:

Option 3

Step-by-step explanation:

Since the denominators are the same, you can just subtract the numerators.

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Garica family bought 9 bags of cookies each bag had 15 cookies they have eaten 27 of the cookies how many cookies do they have l

vova2212 [387]

108 because 9 x 15 is 135 then 135- 27 is 108

3 0

3 years ago

What is the absolute value of any nonzero numberraised to the power of zero

jarptica [38.1K]

The answer to your question is 0

8 0

3 years ago

Brainliest for correct answer

il63 [147K]

Answer:

V =1884 in ^3

Step-by-step explanation:

The volume of a cylinder is given by

V = pi r^2 h where r is the radius and h is the height

V = pi ( 10) ^2 * 6

V = pi ( 600) in^3

Using 3.14 for pi

V = 3.14 * 600 in ^3

V =1884 in ^3

3 0

2 years ago

The scale factor from M to L is 3/2. What is the scale factor from L to M?

spin [16.1K]

Answer:

2/3

Step-by-step explanation:

its right

6 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

3 years ago