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3 years ago

⚠️THIS IS DUE TODAY⚠️ USE DISTRIBUTIVE PROPERTY

Mathematics

1 answer:

Iteru [2.4K]3 years ago

7 0

Answer:

4x + 8

-45 - 5x

7x - 21

Step-by-step explanation:

For the first one

4(x + 2)

for distributive property you multiply the number or symbol next to the outside of the parenthesis, by everything inside the parenthesis.

4(x+2)

4*x = 4x

4*2 = 8

So, this is 4x + 8

-5(9+x)

-5*9= -45

-5*x=-5x

So, this is -45 - 5x

7(x-3)

7*x= 7x

7*-3= -21

This will be 7x - 21

Hope this helps!!

-Ketifa

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

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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

$F_1 = -5\sqrt3 v_0$

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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,

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

$= = F$

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Step-by-step explanation:

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