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3 years ago

Can someone pls answer this !!!

Mathematics

1 answer:

damaskus [11]3 years ago

7 0

Answer:

Step-by-step explanation:

Supplementary angles- angles that equal 180

AOC is 44+56

AOC=100

EOB clearly is more than 100 degrees seeing as its wider than AOC so a is not true

Complementary angles- angles that equal 90

AOC is 100 degrees and BOC is 56

thats more than 90 so thats not true either.

Adjacent angles- angles with a common side and vertex and don't overlap. They do not need to have the same measure or add up to anything.

Both EOD and DOC share a vertex, O, and a side, OD, showing they are adjacent making the answer true.

The answer would be C. no, no, yes

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During a period of 112 minutes, a music station played 40 minutes of commercials. What

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Answer:

9:5

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3 years ago

Find the center of mass of the wire that lies along the curve r and has density =4(1 sin4tcos4t)

dolphi86 [110]

The mass of the wire is found to be 40π√2 units.

To calculate the mass of the wire which runs along the curve r ( t ) with the density function δ=5.

The general formula is,

Mass = $\int_a^b \delta\left|r^{\prime}(t)\right| d t$

To find, we must differentiate this same given curve r ( t ) with respect to t to estimate |r'(t)|.

The given integration limits in this case are a = 0, b = 2π.

Now, as per the question;

The equation of the curve is given as;

r(t) = (4cost)i + (4sint)j + 4tk

Now, differentiate this same given curve r ( t ) with respect to t.

$\begin{aligned}\left|r^{\prime}(t)\right| &=\sqrt{(-4 \sin t)^2+(4 \cos t)^2+4^2} \\&=\sqrt{16 \sin ^2 t+16 \cos ^2 t+16} \\&=\sqrt{16\left(\sin t^2+\cos ^2 t\right)+16}\end{aligned}$

Further simplifying;

$\begin{aligned}&=\sqrt{16(1)+16} \\&=\sqrt{16+16} \\&=\sqrt{32} \\\left|r^{\prime}(t)\right| &=4 \sqrt{2}\end{aligned}$

Now, use integration to find the mass of the wire;

$\begin{aligned}&=\int_a^b \delta\left|r^{\prime}(t)\right| d t \\&=\int_0^{2 \pi} 54 \sqrt{2} d t \\&=20 \sqrt{2} \int_0^{2 \pi} d t \\&=20 \sqrt{2}[t]_0^{2 \pi} \\&=20 \sqrt{2}[2 \pi-0] \\&=40 \pi \sqrt{2}\end{aligned}$