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3 years ago

One angle of a triangle measures 20°. the other two angles are in a ratio of 5:11. what are the measures of those two angles?

Mathematics

1 answer:

Andreyy893 years ago

5 0

They are 50 degrees and 110 degrees. Hope it helps!

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What is the slope of this line?

Talja [164]

Answer:

-1

Step-by-step explanation:

Slope= rise/run

Since the line is decreasing, it will be negative.

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2 years ago

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alexira [117]

24 pine trees in equal rows, with the number of trees in each row being 10 or more than the number of rows.
Hugo could have planted 12 trees in 2 rows.

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3 years ago

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Which of the following is equivalent to the complex number i^13

mestny [16]

Answer:i

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3 years ago

If a person walks 1/6 mile in 10 mintues how far will that person walk in one hour

zvonat [6]

Answer:

1 mile

Step-by-step explanation:

We can write a ratio to solve

We know 1 hour is 60 minutes

1/6 mile x miles

------------ = ---------------

10 minutes 60 minutes

Using cross products

1/6 miles * 60 minutes = 10 minutes * x

10 = 10x

Divide by 10

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1 mile

8 0

2 years ago

Find the area of the surface. the part of the plane 4x + 4y + z = 16 that lies inside the cylinder x2 + y2 = 9

MAVERICK [17]

As a first step, convert to cylindrical coordinates. The equation $x^2+y^2=9=3^2$ is a clue to set

$x=3r\cos\theta$

$y=3r\sin\theta$

then have $0\le r\le1$ and $0\le\theta\le2\pi$ . Meanwhile, the plane equation tells you to use

$4x+4y+z=16\implies z=16-12r\cos\theta-12r\sin\theta$

So we parameterize the part of the plane within the cylinder with the vector-valued function

$\mathbf s(r,\theta)=(3r\cos\theta,3r\sin\theta,15-12r\cos\theta-12r\sin\theta)$

The surface integral giving the area of the surface is

$\displaystyle\iint_{\mathcal S}\mathrm dS$

where $\mathcal S$ denotes the surface in question, and the surface element $\mathrm dS$ is

$\mathrm dS=\left\|\dfrac{\partial\mathbf s}{\partial r}\times\dfrac{\partial\mathbf s}{\partial\theta}\right\|\,\mathrm dr\,\mathrm d\theta$

We have

$\dfrac{\partial\mathbf s}{\partial r}=(3\cos\theta,3\sin\theta,-12(\cos\theta+\sin\theta))$

$\dfrac{\partial\mathbf s}{\partial\theta}=(-3r\sin\theta,3r\cos\theta,12r(\sin\theta-\cos\theta))$

$\implies\mathrm dS=\|(36r,36r,9r\|\,\mathrm dr\,\mathrm d\theta=9\sqrt{33}r\,\mathrm dr\,\mathrm d\theta$

So the area is

$\displaystyle\iint_{\mathcal S}\mathrm dS=9\sqrt{33}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta=18\sqrt{33}\,\pi$

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3 years ago