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2 years ago

Can someone help me answer these!

Mathematics

1 answer:

Marysya12 [62]2 years ago

3 0

Answer:

I believe it would most likely be the last answer

Step-by-step explanation:

I am not sure what it is showing though. All i see is that it is emphisising the multiplication property with the use of perenthisis.

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You have conducted a survey on a number of households and found out how many pets they owned brainly

eduard

Answer:

Using tally marks

Step-by-step explanation:

Missing information;

Number of pets are;

3,0,1,4,4,1,2,0,2,2,0,2,0,1,3,1,2,1,1,3

Find;

Ungroups frequency distribution table

Computation:

<u>Number of pets Tally Frequency</u>

0 IIII 4

1 IIIII I 6

2 IIIII 5

3 III 3

4 II 2

4 0

3 years ago

A single-engine plane is flying at 30 mph in the direction N 60° E. A wind is blowing from west to east at 15 mph

Ksivusya [100]

Answer:

C on Edge 2020

Step-by-step explanation:

The graph where the 60 degree angle is touching the y axis and air is above ground

7 0

3 years ago

Read 2 more answers

Prove that: u+v+w=0 = uxv = vxw = wxu

Misha Larkins [42]

Posting screenshot of my answer in the linked question in case it gets deleted.

4 0

3 years ago

The area of a circle is 161 in. What is the radius?

kirill [66]

Answer:

Approximately 7.16 in

Step-by-step explanation:

Hi there!

Area of a circle equation: $A=\pi r^2$ where r is the radius

Plug in the given area 161 in.²

$161=\pi r^2$

Divide both sides by π to isolate r²

$\frac{161}{\pi} =\frac{\pi r^2}{\pi} \\\frac{161}{\pi} =r^2$

Take the square root of both sides to isolate r

$=r^2\\\sqrt{\frac{161}{\pi}} =r\\7.16=r$

Therefore, the radius of the circle is approximately 7.16 in.

I hope this helps!

7 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