All categories

3 years ago

A line is a undefined term because

1 answer:

7 0

The best answer I can think of is that a line doesn't have a point where it starts or stops, but instead goes on forever, unlike a ray which has a start or end point but not the other, and a line segment which has a start and stop point. Hope this helps!

You might be interested in

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

3 years ago

I GIVE BRAINLIEST PLS HELP

ioda

The answer is 8
Here's why:
${ ( \frac{( {6}^{7}) \times ( {3}^{3}) }{( {6}^{6}) \times ( {3}^{4} ) } )}^{3} = \\ ( \frac{6}{3} ) ^{3} = \\ \frac{216}{27} = 8$
The exponents are subtracted one from another when divided.
$\frac{ a ^{b} }{ {a}^{c} } = {a}^{b - c}$
We can look at the problem this way:
$( \frac{6^{7} }{6 ^{6} } \times \frac{3^{3} }{ {3}^{4} } ) = (6^{7 - 6} \times {3}^{3 - 4} ) = \\ ({6}^{1} \times {3}^{ - 1} )$
Since we have the power of -1 on the 3, we apply this rule:
${a}^{ - b} = \frac{1}{ {a}^{b} }$
Also this rule because we have the power of 1 on the 6:
${a}^{1} = a$
Then we get this:
$(6 \times \frac{1}{3} )^{3} = ( \frac{6}{3} )^{3}$
We apply the rule:
$( \frac{a}{b}) ^{c} = \frac{ {a}^{c} }{ {b}^{c} }$
We get this:
$\frac{{6}^{3} }{ {3}^{3} } = \frac{216}{27} = 8$

4 0

4 years ago

Suppose that 24 inches of wire costs 96 cents.

Alexxx [7]

At the same rate, 31 inches of wire will cost 124 cents.

3 0

2 years ago

Solve for y -3y-12=-4

Brilliant_brown [7]

Answer:

−8/3

Step-by-step explanation:

Step 1: Add 12 to both sides.

−3y−12+12=−4+12

−3y=8

Step 2: Divide both sides by -3.

−3y−3=8−3

y=−8/3

8 0

3 years ago

A homeowner chooses 9 of 25 plants at a nursery to landscape her front yard. Which expression gives her the number of ways she c

Vladimir79 [104]

The correct answer is 25C9

6 0

3 years ago