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3 years ago

If m is perpendicular to n, solve for x

Mathematics

2 answers:

bixtya [17]3 years ago

6 0

x + 21 + 3x + 9 = 90

Combine like terms

4x + 30 = 90

Subtract 30 from both sides

4x = 60

Divide both sides by 4

x = 15

aalyn [17]3 years ago

5 0

Answer:

x = 15

Step-by-step explanation:

You can use the opposite angles property to solve this problem. By using the property, the opposite angle of x+21 must also be x+21. So therefore, you can setup the equation x+21+3x+9=90. Simplify and you should get 4x+30=90. Subtract 30 on both sides and you get 4x=60. Divide 4 on both sides and you get x=15.

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Earnings per hour and greatest hourly rate

horrorfan [7]

This can be solved by using a simple algebraic equation.

For example, in the first column, Caleb worked 5 hours. We do not know how much he makes per hour, so let that be x. He earned a total of $36.25, so the equation can be written as 5x=36.25.

Divide by 5 on both sides, and you get x= 7.25

He earns $7.25 per hour of work.

Repeat this for all the workers:

Jeremy: 7.5x=65.25

x= $8.70

Maria: 4.25x=34.00

x= $8

Rosa: 8x=54.00

x= $6.75

Jeremy has the greatest hourly rate

5 0

4 years ago

Read 2 more answers

A yo-yo is moving up and down a string so that its velocity at time t is given by v(t) = 3cos(t) for time t ≥ 0. The initial pos

jeka57 [31]

Part A - The average value of v(t) over the interval (0, π/2) is 6/π

Part B - The displacement of the yo-yo from time t = 0 to time t = π is 0 m

Part C - The total distance the yo-yo travels from time t = 0 to time t = π is 6 m.

<h3>Part A: Find the average value of v(t) on the interval (0, π/2)</h3>

The average value of a function f(t) over the interval (a,b) is

$f(t)_{avg} = \frac{1}{b - a} \int\limits^b_a {f(t)} \, dx$

So, since velocity at time t is given by v(t) = 3cos(t) for time t ≥ 0. Its average value over the interval (0, π/2) is given by

$v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {v(t)} \, dt$

Since v(t) = 3cost, we have

$v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {3cos(t)} \, dt\\= \frac{3}{\frac{\pi }{2}} \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= \frac{6}{{\pi}} [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= \frac{6}{{\pi}} [{sin(\frac{\pi }{2})} - sin0]\\ = \frac{6}{{\pi}} [1 - 0]\\ = \frac{6}{{\pi}} [1]\\ = \frac{6}{{\pi}}$

So, the average value of v(t) over the interval (0, π/2) is 6/π

<h3>Part B: What is the displacement of the yo-yo from time t = 0 to time t = π?</h3>

To find the displacement of the yo-yo, we need to find its position.

So, its position x = ∫v(t)dt

= ∫3cos(t)dt

= 3∫cos(t)dt

= 3sint + C

Given that at t = 0, x = 3. so

x = 3sint + C

3 = 3sin0 + C

3 = 0 + C

C = 3

So, x(t) = 3sint + 3

So, its displacement from time t = 0 to time t = π is

Δx = x(π) - x(0)

= 3sinπ + 3 - (3sin0 + 3)

= 3 × 0 + 3 - 0 - 3

= 0 + 3 - 3

= 0 + 0

= 0 m

So, the displacement of the yo-yo from time t = 0 to time t = π is 0 m

<h3>Part C: Find the total distance the yo-yo travels from time t = 0 to time t = π. (10 points)</h3>

The total distance the yo-yo travels from time t = 0 to time t = π is given by

$x(t) = \int\limits^{\pi}_0 {v(t)} \, dt\\= \int\limits^{\pi }_0 {3cos(t)} \, dt\\= 3 \int\limits^{\pi }_0 {cos(t)} \, dt\\ = 3 \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt + 3\int\limits^{\pi }_{\frac{\pi }{2}} {cos(t)} \, dt\\= 3 \times 2\int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= 6 [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= 6[{sin\frac{\pi }{2} - sin0]\\\\= 6[1 - 0]\\= 6(1)\\= 6$