Find u and v .

I need help

[tex]Find u and v .... I need help[/tex]

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Find u and v .

I need help

[tex]Find u and v .... I need help[/tex]

step-by-step explanation: a reaction is first order. if the initial reactant concentration is 0.0200 m, and 25.0 days later the concentration is 6.25 x 10-4 m, then its half-life is:

hello your question is incomplete attached below is the complete question

A) xy ≥ 0, (cx )(cy) = c^2(xy) ≥ 0

B) U + V is not in W

Step-by-step explanation:

attached below is the detailed solution

A) since xy ≥ 0, (cx )(cy) = c^2(xy) ≥ 0

B ) U + V is not in W

[tex]Let W be the union of the and quadrants in the xy-plane. That is, let . Complete parts a and b bel[/tex]

[tex]Let W be the union of the and quadrants in the xy-plane. That is, let . Complete parts a and b bel[/tex]

1. Find a ⋅ b.

a = <7, 4>, b = <3, 5> (2 points)

41 < right answer

<10, 9>

-1

<21, 20>

Explanation:

a . b means the dot or scalar product of the vectors a and b.

The scalar product of vectorr <x1,y2> and <x2,y2> is (x1)*(y1) + (x2)*(y2)

So, a . b = 7*3 + 4*5 = 21 + 20 = 41 < answer

2. Find a ⋅ b

a = 10i + 9j, b = 4i + 3j (2 points)

<40, 27>

<14, 12>

67 < answer

-13

Explanation:

Again you are required to find the dot product of two vectors. In this case the vectors are given using the representation with the unit vectors i and j.

The dot product of the vectors (x1 i + y1 j) . (x2 i + y2 j) is x1*x2 + y1*y2

So, (10i + 9j) . (4i + 3j) = 10*4 + 9*3 = 40 + 27 = 67

67

3. Find the angle between the given vectors to the nearest tenth of a degree.

u = <6, -1>, v = <7, -4> (2 points)

20.3° < answer

10.2°

0.2°

30.3°

Explanation:

You can use the dot product to find the angle between two vectors.

This is the formula:

cos(α) = [ <a> dot <b. ] / [|a| |b| ]

where α is the angle between the vector a and b.

<a> is the vector a, <b> is the vector b, <a> dot <b> is the dot product, |a| is the magnitude of vector a, and |b| is the magnitude of vector b.

<u> dot <v> = (6)(7) + (-1)(-4) = 42 + 4 = 46

|u| = √ (6^2 + 1^2) = √37

|v| = √ (7^2 + 4^2) = √65

=> cos(α) = [46] / [√37 * √65] = 0.938

=> α = arccos(0.938) = 20.3°

4. Determine whether the vectors u and v are parallel, orthogonal, or neither.

u = <10, 6>, v = <9, 5> (2 points)

Orthogonal

Parallel

Neither < answer

Explanation:

At sight they are neither orthogonal nor parallel. You can put the points in a graph and you will realize inmediately.

Let's calculate the angle, with the formula given in the above problem.

cos(α) = [<u> dot< >v] / [ |u| |v| ]

<u> dot <v> = 10*9 + 6*5 = 90 + 30 = 120

|u| = √(10^2 + 6^2) = √136

|v| = √(9^2 + 5^2) = √106

cos(α) = 120 / (√136 * √106] = 0,999

α = arctan(0,999) = 1,9°

Which confirms that they are neither orthogonal nor parallel

5. Evaluate the expression:

v ⋅ w

Given the vectors:

r = <8, 8, -6>; v = <3, -8, -3>; w = <-4, -2, -6> (2 points)

Solution:

v . w = v dot w = (3)(-4) + (-8)(-2) + (-3)(-6) = -12 + 16 + 18 = 22

22

the solution is 12x-6y

26 ,32.2

Step-by-step explanation:

on Edge <3

A: P=Po(Vo/V)^y

B: W=5/2Po(Vo-Vo^7/5 Vi^-2/5)

C: ∆V= 5/2Po[Vo^7/5 Vi^-2/5 -Vo]

Explanation:

Attached is the solution

[tex]In this problem you are to consider an adiabaticexpansion of an ideal diatomic gas, which means that[/tex]

[tex]In this problem you are to consider an adiabaticexpansion of an ideal diatomic gas, which means that[/tex]

The magnitude of w is approximately 26.

The angle between w and u is approximately 32.2°.

Step-by-step explanation:

Consider the travel of the speedboat from A to B and back.

Refer to the first figure.

The boat travels at 5 m/s relative to the water, and the downstream speed of the water is 0.5 m/s.

Therefore,

V₁=5 m/s, u = 0.5 m/s, sinθ = 0.5/5 = 0.1 => θ = arcsin(0.1) = 5.74°.

The boat should travel upstream at 5 m/s, at an angle of 5.74°.

Similarly, return speed from B to A is 5 m/s, at 5.74° upstream.

The horizontal component of velocity from A to B (or vice versa) is

5cos(5.74°) = 4.975 m/s.

The time required to travel 1000 m is

1000/4.975 = 202.02 = 3.367 min.

The time for the return trip is

t₁ = 2*3.367 = 6.734 min.

Consider travel from C to D and back, as shown in the second figure.

The resultant velocity upstream from C to D is

V₁ = 5 - 0.5 = 4.5 m/s

The time required to travel 1000 m fromC to D is

1000/4.5 = 222.22 s

The resultant velocity downstream from D to C is

V₂ = 5 + 0.5 = 5.5 m/s

The time required to travel fro D to C is

1000/5.5 = 181.82 s

Total time for the return trip between C and D is

t₂ = 222.22 + 181.82 = 404.04 s = 6.734 min

The first boat should travel upstream at an angle of 5.74°. The time for the return trip between A and B is t₁ = 6.734 min or 404.04 s.

The second boat travels upstream against the current, and downstream with the current. The time for the return trip between C and D is t₂ = 6.734 min or 404.04 s.

[tex]Two speedboats can each travel at v= 5 m/s with respect to water. one boat crosses the river of the[/tex]

[tex]Two speedboats can each travel at v= 5 m/s with respect to water. one boat crosses the river of the[/tex]

a)

The analytic expression for p(V) is [tex]P(V) = P_{0}(\frac{V_{0}}{V})^{7/5}[/tex]

b)

The expression of the work in terms of p0 ,V0,V1 is

[tex]W= \frac{5}{2} P_{0}[V_{0}-V_{0}^{7/5}V_{1}^{-2/5}][/tex]

c)

The change in internal energy of the gas

[tex]\delta U = \frac{5}{2}P_{0}[V_{0}^{7/5}V_{1}^{-2/5}-V_{0}][/tex]

Explanation:

The explanation is shown on the first, second and third

[tex]In this problem you are to consider an adiabaticexpansion of an ideal diatomic gas, which means that[/tex]

[tex]In this problem you are to consider an adiabaticexpansion of an ideal diatomic gas, which means that[/tex]

[tex]In this problem you are to consider an adiabaticexpansion of an ideal diatomic gas, which means that[/tex]

U is up in top left v is in bottom left

21.5°

option D is the correct option.

Please see the attached picture for full solution..

Hope it helps..

Good luck on your assignment.

[tex]I NEED HELP PLEASE, THANKS! Find the angle θ between u and v if u = <–9, –8 , 9> and v = <[/tex]

Given:

u and v are parallel lines.

To find:

The value of x.

Solution:

If a transversal line intersect the two parallel lines, then

(1) Corresponding angles are congruent.

(2) Same sides interior angles are supplementary angles and their sum is 180 degrees.

15.

[tex]23x-2=90[/tex] (Corresponding angles)

[tex]23x=90+2[/tex]

[tex]x=\dfrac{92}{23}[/tex]

[tex]x=4[/tex]

Therefore, the value of x is 4.

16.

[tex](x+106)+(x+86)=180[/tex] (Same sided interior angles)

[tex]2x+192=180[/tex]

[tex]2x=180-192[/tex]

[tex]2x=-12[/tex]

Divide both sides by 2.

[tex]x=\dfrac{-12}{2}[/tex]

[tex]x=-6[/tex]

Therefore, the value of x is -6.