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4 years ago

I need helpPlease and Fast! What is the distance between (-3, -2) and (5, 4)?

Mathematics

1 answer:

nexus9112 [7]4 years ago

7 0

Using the Pythagorean Theorem, we can tell that there is 10 units between the two points.

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50 points :] !!

nasty-shy [4]

Given,

LP = 15, PR = 9

Point P lies on the line segment PR. It would mean that,

LP + PR = LR

⇒LR = 15 + 9

⇒ LR = 24

Hence, "LR = 24 because LP + PR = LR according to the Segment Addition Postulate, and 15 + 9 = 24 using substitution" is the correct option.

6 0

3 years ago

Leigh have a piece of rope that is 6 2/3 feet long. How do you write 6 2/3 as a fraction greater than 1?

Fed [463]

20/3 or 20 over three i think

4 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

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