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3 years ago

13

Please helppppp!!!!!!!

2 answers:

ale4655 [162]3 years ago

8 0

Answer:

K 2

J 4

Step-by-step explanation:

Solved it as an equation 2 x+x=6

x= number of miles Katie drives

Sladkaya [172]3 years ago

3 0

It is 12 Multiplication

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3 years ago

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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

I Need The Answer Plz Geometry Is Hard!!

otez555 [7]

Answer:

14

Step-by-step explanation:

In a triangle, the midline joining the midpoints of two sides is parallel to the third side and half as long, so

FB = EC/2

2x - 6 = 44/2

2x - 6 = 22

2x = 22 + 6

2x = 28

x = 28/2

x = 14

6 0

3 years ago

Can u help me in this <br> plz plz

e-lub [12.9K]

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7 0

3 years ago