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3 years ago

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The terminal side of an angle θ in standard position passes through the point (2,7). calculate the values of the six trigonometr

ic functions for angle θ.

1 answer:

Ksivusya [100]3 years ago

8 0

Brainly doesn't support all the characters put in so I had to resort to a different program and then send it in as an attachment

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Please help me with this homework

Goshia [24]

Every time the line is vertical, the slope is ‘no slope or undefined’.

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2 years ago

Read 2 more answers

Simplify to create an equivalent expression

Keith_Richards [23]

B 33k- 15 hope it helps

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3 years ago

Find the area of the following shape. You must show all work to receive credit

kirza4 [7]

Answer:

42 square units

Step-by-step explanation:

Draw segments - 5 to A, A to the point of intersection of Band C and from from the point of intersection of B, C upto 2. It will form a square with each side 8 units and four triangles.

Subtracting the area of 4 four triangles from the area of the square will give you the area of the required shape.

Area of the shape

$= {8}^{2} - \frac{1}{2} \times 3 \times 6 - \frac{1}{2} \times 5 \times 2 - \frac{1}{2} \times 2 \times 3\\ - \frac{1}{2} \times 2 \times 5 \\ \\ = 64 - 3 \times 3 - 5 \times 1 - 1\times 3 - 1 \times 5 \\ \\ = 64 - 9 - 5 - 3 - 5 \\ \\ = 64 - 22 \\ = 42 \: {units}^{2}$

3 0

2 years ago

Read 2 more answers

The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo

oksano4ka [1.4K]

Answer:

Required center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

Step-by-step explanation:

Given semcircles are,

$y=\sqrt{1-x^2}, y=\sqrt{16-x^2}$ whose radious are 1 and 4 respectively.

To find center of mass, $(\bar{x},\bar{y})$ , let density at any point is $\rho$ and distance from the origin is r be such that,

$\rho=\frac{k}{r}$ where k is a constant.

Mass of the lamina=m= $\int\int_{D}\rho dA$ where A is the total region and D is curves.

then,

$m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k$

Now, x-coordinate of center of mass is $\bar{y}=\frac{M_x}{m}$ . in polar coordinate $y=r\sin\theta$

$\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta$

$=3k\int_{0}^{\pi}\sin\theta d\theta$

$=3k\big[-\cos\theta\big]_{0}^{\pi}$

$=3k\big[-\cos\pi+\cos 0\big]$

$=6k$

Then, $\bar{y}=\frac{M_x}{m}=\frac{2}{\pi}$

y-coordinate of center of mass is $\bar{x}=\frac{M_y}{m}$ . in polar coordinate $x=r\cos\theta$

$\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta$

$=3k\int_{0}^{\pi}\cos\theta d\theta$

$=3k\big[\sin\theta\big]_{0}^{\pi}$

$=3k\big[\sin\pi-\sin 0\big]$

$=0$

Then, $\bar{x}=\frac{M_y}{m}=0$

Hence center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

3 0

2 years ago

What is this asking? 8x-4y=12

kkurt [141]

Maybe from the equation of a line from y=mx+c
Where c ( intercept) is -12

8 0

2 years ago