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3 years ago

Pls help meeethank you thank you thank you thank you

Mathematics

2 answers:

olga_2 [115]3 years ago

6 0

Answer:

See below

Step-by-step explanation:

11 box 1: 12

11 box 2: 101

12 box 1: 80

12 box 2: 80

12 box 3: 80

13 box 1: 16

13 box 2: 16

13 box 3: 76

20 = y + 12

20 - 12 = y + 12 - 12

8 = y

x + 2 = 19

x + 2 - 2 = 19 - 2

x = 17

z - 313 = 176

z - 313 + 313 = 176 + 313

z = 489

natka813 [3]3 years ago

5 0

Answer:

See below

Step-by-step explanation:

11 box 1: 12

11 box 2: 101

12 box 1: 80

12 box 2: 80

12 box 3: 80

13 box 1: 16

13 box 2: 16

13 box 3: 76

20 = y + 12

20 - 12 = y + 12 - 12

8 = y

x + 2 = 19

x + 2 - 2 = 19 - 2

x = 17

z - 313 = 176

z - 313 + 313 = 176 + 313

z = 489

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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

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 the sum of the difference?

Alona [7]

Answer:

A. 9x^4 and 3x^5y

Step-by-step explanation:

there are two ways to solve this:

first way:

You can solve this my substituting numbers for x and y in this case i used 2 for x and 3 for y and see which one is equal to the original equations

the second way is the regular way

when you add or subtract numbers with variables and exponents you want to add the constants and add the exponents in this case $5x^{4} +4x^2$ is the same as $(5x+4x)^(4+4)$ = $9x^4$ and you can do the same process for subtraction