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1 year ago

Polygon ABCD has vertices A(1,3), B(1,6), C(4,6), and D(5,2)

Mathematics

1 answer:

Galina-37 [17]1 year ago

7 0

Answer: A (0.5, 1.5) B(0.5, 3) C(2, 3) D(2.5, 1)

Step-by-step explanation:

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49°<br> Solve for 42.<br> 2 = [?]<br> 44/62

Brrunno [24]

Answer:

∠2 = 41°

Step-by-step explanation:

∠2 + 49° + 90° = 180°

∠2 + 139° = 180°

∠2 = 180° - 139°

∠2 = 41°

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2 years ago

Read 2 more answers

If Tom brushes his teeth 1/12 hour a day. how many minutes does Tom brush his teeth in 2 weeks? What units of time do you need t

lilavasa [31]

1/12 hour = 5 minutes, 2 weeks is 14 days, so 5*15 = the amount of time he brushes his teeth so you would need to change from hours to minutes

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2 years ago

Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions

quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of $v_0$ and $F_2$ be the component normal to $v_0$ .

Since, $|v_0| = 1,$

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, $v_0 = \left$

From figure,

$|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3$

We know that the direction of $F_1$ is opposite of the direction of $v_0$ , so we have

$F_1 = -5\sqrt3 v_0$

$=-5\sqrt3 \left$

$= \left$

The unit vector in the direction normal to the plane, $v_1$ has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

$|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5$

∴ $F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>$

$=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

$= = F$

3 0

2 years ago

What is –36° converted to radians?

Stells [14]

let's recall that there are 180° in π radians, thus

$\bf \begin{array}{ccll} degrees&radians\\ \cline{1-2} 180&\pi \\ -36&x \end{array}\implies \cfrac{180}{-36}=\cfrac{\pi }{x}\implies -5=\cfrac{\pi }{x}\implies x=\cfrac{\pi }{-5}\implies x=-\cfrac{\pi }{5}$