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saw5 [17]
3 years ago
12

Find the value of x and y

Mathematics
1 answer:
AVprozaik [17]3 years ago
5 0

Answer:

x = 10 and y = 15

Step-by-step explanation:

The arrows on lines AB and CD indicate that they are parallel. This means that we have to look for Z-angles (I don't know if they are called the same where you're from but I hope you recognise the term). This means that ∠ABC = ∠BCD

5x - 20° = x + 20°

4x = 40°

x = 10°, which gives us the first answer.

The angle formed by points ACE (a straight line) is 180 degrees. Lets refer to this as angle ACE. We also know this angle consists of angle ACB, BCD and DCE. This gives us the following equation:

angle ACE = angle ACB + x + 20° + y + 20° = 180°

angle ACB + x + y + 40° = 180°

angle ACB = -x - y + 140°

Inserting the x we found earlier this gives

∠ACB = - y + 130°

Now we can find a second equation to describe ACB by looking at the triangle it is inside. The angles in a triangle always add up to 180°, giving the following expression:

4y - 25° + 5x - 20° + ∠ACB = 180°

Substituting x again gives

4y - 25° + 50° - 20° + ACB = 180°

ACB = 180 - 4y - 5° = 175° -4y

Combining the two expressions for ACB gives the final equation to solve:

175° - 4y = -y + 130°

-3y = -45

y = 15

If you have any questions still feel free to reach out!

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Find the magnitude and the positive direction angle for u. u = (-12, 5) theta = degree (Round to the nearest tenth as needed.) F
Ivan

Question:

1. Find the magnitude and the positive direction angle for u.

u = (-12, 5)

For theta degree, round to the nearest tenth as needed.

2. For the vector r = (-9, -10), find -9r.

Answer:

(1)

The magnitude of vector u is 13.

The positive direction of vector u is 157.4° (to the nearest tenth)

(2)

-9r = (81, 90)

Step-by-step explanation:

(1) For a given vector u = (a, b), the magnitude of u is given by;

|u| = \sqrt{a^2 + b^2}            --------------------(i)

The direction of such vector is given by;

θ = tan ⁻¹ (\frac{b}{a})                    -------------(ii)

Where;

a and b are the x and y components of the vector.

<em>Now,</em>

From the question,

u = (-12, 5)

(a) From equation (i), the magnitude of u is therefore;

|u| = \sqrt{(-12)^2 + (5)^2}

|u| = \sqrt{144 + 25}

|u| = \sqrt{169}

|u| = 14

Therefore, the magnitude of vector u is 13.

(b) From equation (ii) the direction of vector u is therefore;

θ = tan ⁻¹ (\frac{5}{-12})

θ = tan ⁻¹ (-0.41667)

θ = -22.62°

<em>The direction here is negative, but we have been told to find the positive direction.</em>

From the given x and y components of the vector, it can be deduced that the the vector lies in the negative x direction and the positive y direction.

Therefore, the vector lies in the second quadrant. This is shown in the diagram attached to this response.

To get the positive direction, m, we need to add 180° to the result of θ = -22.62° i.e

m = 180° + (-22.62)°

m = 157.38°

Therefore, the positive direction of vector u is 157.4° (to the nearest tenth)

(2) For the vector r = (-9, -10), we are asked to calculate -9r.

Since;

r = (-9, -10)

-9r = -9(-9, -10)    [Multiply by -9]

-9r = (81, 90)

Therefore, -9r = (81, 90)

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Answer: the second answer 5p

Step-by-step explanation:

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