PLEASE HELP ME I AM SUFFERING GREATLY

Solution :

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

Since, [tex]|v_0| = 1,[/tex]

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

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

Therefore, [tex]v_0 = \left<\frac{1}{2},\frac{\sqrt 3}{2}\right>[/tex]

From figure,

[tex]|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3[/tex]

We know that the direction of [tex]F_1[/tex] is opposite of the direction of [tex]v_0[/tex], so we have

[tex]$F_1 = -5\sqrt3 v_0$[/tex]

    [tex]$=-5\sqrt3 \left<\frac{1}{2},\frac{\sqrt3}{2} \right>$[/tex]

    [tex]$= \left<-\frac{5 \sqrt3}{2},-\frac{15}{2} \right>$[/tex]

The unit vector in the direction normal to the plane, [tex]v_1[/tex] has components :

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

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

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

From figure,

[tex]|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5[/tex]

∴  [tex]F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>[/tex]

                   [tex]$=\left<\frac{5 \sqrt3}{2},-\frac{5}{2} \right>$[/tex]

Therefore,

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

           [tex]$=<0,- 10> = F$[/tex]

Answer:

625 N

Step-by-step explanation:

First, find the area of the square.

Area = side x side

        = 5 x 5

        = 25 m²

Force = pressure x area

          = 25 x 25

          = 625 N

Hope this helps!

Step-by-step explanation:

first number=x

second number=x+2

third number=x+4

(x+2)(x+4)=35

x(x+4)+2(x+4)=35

x²+4x+2x+8=35

x²+6x+8=35

x²+6x=35-8

x²+6x=33

x²+6x-33=0

using that, you can find the first number,

then use the data to find the other two.

Answer:

There is no common factor

Step-by-step explanation:

Answer:

[tex]\sqrt{15}[/tex]

Step-by-step explanation:

Using the rule of radicals

[tex]\sqrt{a}[/tex] × [tex]\sqrt{b}[/tex] ⇔ [tex]\sqrt{ab}[/tex] , then

[tex]\sqrt{3}[/tex] × [tex]\sqrt{5}[/tex] = [tex]\sqrt{3(5)}[/tex] = [tex]\sqrt{15}[/tex]

Answer:

The value of x is 5

Answer:

1/ 4^2

Step-by-step explanation:

4^4 * 4^-6

We know that a^b * a^c = a^(b+c)

4^(4-6)

4^-2

We also know that a^-b = 1/a^b

1/4^2

Answer:

6 sides

Step-by-step explanation:

The interior angle and the exterior angle sum to 180° , then

exterior angle = 180° - 120° = 60°

The sum of the exterior angles of a polygon is 360° , so

number of sides = 360° ÷ 60 = 6

Answer:

6 sides

Step-by-step explanation:

180 - 120 =60°

sum of exterior angles is 360°

therefore , 360°/60°

ans = 6 sides

Good luck :-)

Answer:

2(6.3) + b = 15.7

12.6 + b = 15.7

b = 3.1

Answer:

y = (x - 3)² - 4      Vertex form

(3, -4)    Vertex

Step-by-step explanation:

f(x) = x² - 6x + 5

Complete the square

y = (x - 3)² + 5 - (-3)²

y = (x - 3)² + 5 - 9

y = (x - 3)² - 4      Vertex form

(3, -4)    Vertex

Answer:

Circumference of circle is 9.42cm.

Step-by-step explanation:

Here we have given that A circle with having radius 3 and 1 / 2 cm. And We need to find circumference of circle.

Formula using

Circumference of circle = π × 2 × radius

substitute the value

Circumference of circle = π × 2 × 3 and 1/2 cm.

cancle the 2

Circumference of circle = π × 3 cm

Using π = 22 / 7

[tex] \small \sf \ Circumference \: of \: \: circle = \frac{ 22}{7 }×3 cm \\ [/tex]

[tex] \small \sf \ Circumference \: of \:circle = 9.42 cm[/tex]

Hence, Circumference of circle is 9.42cm.

2

Hope this helps! :)

______________

Answer:

you can know by the minimum number or the maximum e.g minimum 1 maximum 20 answer 20

Step-by-step explanation:

so in this kind it is 2

hope its useful

Answer: 12.

Step-by-step explanation: Use Pythagorean Theorem. We that a = 9, b = ?, and c = 15. Plugging that in gives us 9^2 + b^2 + 15^2. Solving for b gets us b^2 = 144. Taking the square root of both sides gets us b = 12.

Answer:

[tex] {8x}^{3} + 27 \\ = {8x}^{3} + {3}^{3} \\ let : 8x \: be \: a \\ : 3 \: be \: b \\ = > {a}^{3} + {b}^{3} : \\ {(a + b)}^{3} = (a + b)( {a}^{2} + 2ab + {b}^{2} ) \\ {(a + b)}^{3} = ( {a}^{3} + 3{a}^{2} b + 3a {b}^{2} + {b}^{3} ) \\ ( {a}^{3} + {b}^{3} ) = {(a + b)}^{3} - 3ab(a + b) \\ \therefore( {8x}^{3} + 27) = {(8x + 3)}^{3} - 72(8x + 3)[/tex]

Answer:

a+k ≤ 35

8a+4k ≤ 190

Some possible combinations are 10 adults and 20 kids

5 adults and 29 kids

Step-by-step explanation:

Let k= number of kids

a = number of adults

a+k ≤ 35  since you can have at most 35

8a + 4k ≤ 190 since the most you can spend is 190

Some possible combinations are 10 adults and 20 kids

(10+30 < 35 and 8*10+4*20 = 80+80 =160< 190)

Another possible combination

5 adults and 29 kids

(5+29 < 35  and 5*10 + 4*29 =50+116=166)

Answer:

96

Step-by-step explanation:

80*120=9600 divided by 100

Answer:

[tex]\displaystyle y=-\frac{1}{10}x^2-\frac{4}{5}x-\frac{11}{10}[/tex]

Step-by-step explanation:

By definition, any point (x, y) on the parabola is equidistant from the focus and the directrix.

The distance between a point (x, y) on the parabola and the focus can be described using the distance formula:

[tex]d=\sqrt{(x-(-4))^2+(y-(-2))^2[/tex]

Simplify:

[tex]d=\sqrt{(x+4)^2+(y+2)^2}[/tex]

Since the directrix is an equation of y, we will use the y-coordinate. The vertical distance between a point (x, y) on the parabola and the directrix can be described using absolute value:

[tex]d=|y-3|\text{ or } |3-y|[/tex]

The two equations are equivalent. Therefore:

[tex]\sqrt{(x+4)^2+(y+2)^2}=|y-3|[/tex]

Solve for y. We can square both sides. Since anything squared is positive, we can remove the absolute value:

[tex](x+4)^2+(y+2)^2 = (y-3)^2[/tex]

Expand:

[tex](x^2+8x+16)+(y^2+4y+4)=(y^2-6y+9)[/tex]

Isolate:

[tex]x^2+8x+11=-10y[/tex]

Divide both sides by -10. Hence, our equation is:

[tex]\displaystyle y=-\frac{1}{10}x^2-\frac{4}{5}x-\frac{11}{10}[/tex]

Answer:

[tex]43.7[/tex]

Step-by-step explanation:

Since the triangles are similar, the ratio of their corresponding sides is maintained. Therefore, we can up the following proportion to solve for [tex]OP[/tex]:

[tex]\frac{9}{7}=\frac{OP}{34},\\OP=\frac{34\cdot 9}{7}=43.7142857143\approx \boxed{43.7}[/tex]

43.7

Step-by-step explanation:

Since the two are similar triangles, the ratio of the legs of one triangle is equal to the ratio of the legs of the other triangle.

OP/ON = LM/LK

OP/34 = 9/7

Solving for OP,

OP = 34(9/7)

= 43.7

Answer:

117 centimeters

Step-by-step explanation:

Since the longest pieces is 72 centimeters and it’s ratio is 8, that means the individual unit of the ratio is 9 (72/8=9). You can multiply each part of the ratio by 9

2x9=18

3x9=27

We already know the third part of the ratio is 72. Add all of it together to get the original length.

18+27+72=117 cm

Answer:

my answer is in this picture. sorry about my presentation and English

Answer:

the index of y is -2

Step-by-step explanation:

[tex]y^{-2}[/tex]

=[tex]\frac{1}{y^2}[/tex]

no se ingles xd asi que la verdad sorry aun que dime si es potensiacion

1.) 70x/100x^2 = 7/10x

2.) 16x^3/24x = 2x^2/3

3.) 7x(2x+1)/21(2x+1) = x/3

4.) 35x - 35/x^2 - 1 =  35/x + 1

5.) x + 7/x^2 + 4x - 21 = 1/x - 3

6.) x^2 - 6x + 5 = x - 5/5

7.) x^2 - 36/x^2 - 3x - 18 = x + 6/x + 3

8.) x^2 - 3x - 10/x^2 + x - 2 = x - 5/x - 1

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

[tex]\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3}[/tex] (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

[tex](x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2})[/tex] (1b)

Where:

[tex]x, y[/tex] - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

[tex]r_{1}, r_{2}[/tex] - Length of each vector, in kilometers.

[tex]\theta_{1}, \theta_{2}[/tex] - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that [tex]r_{1} = 42\,km[/tex], [tex]r_{2} = 28\,km[/tex], [tex]\theta_{1} = 32^{\circ}[/tex] and [tex]\theta_{2} = 154^{\circ}[/tex], then the resulting coordinates of the final position of the surveyor is:

[tex](x,y) = (42\,km)\cdot (\cos 32^{\circ}, \sin 32^{\circ}) + (28\,km)\cdot (\cos 154^{\circ}, \sin 154^{\circ})[/tex]

[tex](x,y) = (35.618, 22.257) + (-25.166, 12.274)\,[km][/tex]

[tex](x,y) = (10.452, 34.531)\,[km][/tex]

According to this, the resulting vector is locating in the first quadrant. The bearing of the vector is determined by the following definition:

[tex]\theta = \tan^{-1} \frac{10.452\,km}{34.531\,km}[/tex]

[tex]\theta \approx 16.840^{\circ}[/tex]

And the distance from the camp is calculated by the Pythagorean Theorem:

[tex]r = \sqrt{(10.452\,km)^{2}+(34.531\,km)^{2}}[/tex]

[tex]r = 36.078\,km[/tex]

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Answer:

It would be considered stealing.

Step-by-step explanation:

They did willingly give you their card BUT, they trusted you to only spend a certain amount and not go over that certain limit. You went over the limit even though they specifically told you not to go over it so yes, it would be classified as stealing.