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nikklg [1K]
3 years ago
6

Find a vector that is parallel to -9x+2y = 10. Please explain step by step

Mathematics
1 answer:
coldgirl [10]3 years ago
4 0

Answer:


Step-by-step explanation:

Step One: Find the slope of the given Line

2y - 9x = 10                   Add 9x to both sides

2y - 9x + 9x = 9x + 10  Combine

2y = 9x + 10

y = 9/2 x + 5

The parallel line has a slope of 9/2

Step Two: Find the magnitude

sqrt(x^2 + y^2) = r                   You don't have a magnitude. Invent one, r

x^2 + y^2 = r^2                        Both sides have been squared

Step Three: Find the change in y / x

y / x = 9/2

Step Four: Find the value of y

y/x = 9/2                 Multiply both sides by x

y = 9/2 * x

Step Five: Solve for x

x^2 + y^2 = r^2

x^2 +(9/2x)^2 = r^2

x^2 + 81x^2 / 4 = r^2                Change the 1 to common denominator 1 = 4/4

4/4 x^2 + 81x^2/ 4 = r^2

85 x^2/ 4 = r^2                        Multiply both sides by 4/85

4/85 * (85 / 4) x^2 = 4/85 r^2  

x^2 = 4/85 r^2                         Take the square root of both sides

x = +/- 2/sqrt(85) r                   Rationalize the denominator  

x = (+/- 2 sqrt(85) ) / 85 *

Step Six: Solve for y

y = 9/2 x

y = 9/2 * 2 sqrt(85) / 85      Cancel the 2s

y = +/- 9* sqrt(85)/85

Answer

<2*sqrt(85)/85 ,  9 sqrt(85)/85>

You can write the minus answer if you need it. I kind of suspect you won't.

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The weight of the new student is 27 kg.

Average weight

= total weight ÷total number of students

<h3>1) Define variables</h3>

Let the total weight of the 35 students be y kg and the weight of the new student be x kg.

<h3>2) Find the total weight of the 35 students</h3>

<u>45 =  \frac{y}{35}</u>

y= 35(45)

y= 1575 kg

<h3>3) Write an expression for average weight of students after the addition of the new student</h3>

New total number of students

= 35 +1

= 36

Total weight

= total weight of 35 students +weight of new students

= y +x

44.5  =  \frac{y + x}{36}

<h3>4) Substitute the value of y</h3>

44.5 =  \frac{1575 + x}{36}

<h3>5) Solve for x</h3>

36(44.5)= 1575 +x

1602= x +1575

<em>Subtract 1575 from both sides:</em>

x= 1602 -1575

x= 27

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The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes
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Answer:

a)0.08  , b)0.4  , C) i)0.84  , ii)0.56

Step-by-step explanation:

Given data

P(A) =  professor arrives on time

P(A) = 0.8

P(B) =  Student aarive on time

P(B) = 0.6

According to the question A & B are Independent  

P(A∩B) = P(A) . P(B)

Therefore  

{A}' & {B}' is also independent

{A}' = 1-0.8 = 0.2

{B}' = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B')  (only for independent cases)

= 0.2 x 0.4

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Part b)

The probability that the student is late given that the professor is on time

P(\frac{B'}{A}) = \frac{P(B'\cap A)}{P(A)} = \frac{0.4\times 0.8}{0.8} = 0.4

Part c)

Assume the events are not independent

Given Data

P(\frac{{A}'}{{B}'}) = 0.4

=\frac{P({A}'\cap {B}')}{P({B}')} = 0.4

P({A}'\cap {B}') = 0.4 x P({B}')

= 0.4 x 0.4 = 0.16

P({A}'\cap {B}') = 0.16

i)

The probability that at least one of them is on time

P(A\cup B) = 1- P({A}'\cap {B}')  

=  1 - 0.16 = 0.84

ii)The probability that they are both on time

P(A\cap  B) = 1 - P({A}'\cup {B}') = 1 - [P({A}')+P({B}') - P({A}'\cap {B}')]

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

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