how much time would it take for the sound of thunder to travel 1,500 meters. if sound travels at a speed of 330m/sec

Mathematics

1 answer:

omeli [17]3 years ago

5 0

$\frac{330meters}{1second}$ = $\frac{ 1500 meters}{x seconds}$

1500 / 330 = 4.5454...... seconds.

You might be interested in

The local grocery store wants to know if the shoppers prefer Crest or Colgate toothpaste. The store gives all of its shoppers a

Andrej [43]

A, it is a biased ample because for example I may not go there to get toothpaste. Good luck

4 0

4 years ago

Read 2 more answers

Please help, the question below

Licemer1 [7]

Answer:

$\sf R=\left(3, -\dfrac{5}{4}\right)$

Step-by-step explanation:

Given:

P = (1, 5)
Q = (-2, 3)
R = (a, b)
R lies on line x = 3
PR = QR

If point R lies on the line x = 3, then the x-value of point R is 3.

⇒ a = 3

<u />

<u>Distance between two points</u>

$\sf d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

(where (x₁, y₁) and (x₂, y₂) are the two points)

Use the <u>distance formula</u> to derive equations for PR and QR.

Let (x₁, y₁) = P = (1, 5)

Let (x₂, y₂) = R = (3, b)

$\begin{aligned} \sf PR & =\sf \sqrt{(3-1)^2+(b-5)^2}\\ & = \sf \sqrt{4+(b-5)^2} \end{aligned}$

Let (x₁, y₁) = Q = (-2, 3)

Let (x₂, y₂) = R = (3, b)

$\begin{aligned} \sf QR & =\sf \sqrt{(3-(-2))^2+(b-3)^2}\\ & = \sf \sqrt{25+(b-3)^2} \end{aligned}$

As PR = QR, equate the derived equations and solve for b:

$\begin{aligned} \sf PR & = \sf QR \\\sf \sqrt{4+(b-5)^2} & = \sf \sqrt{25+(b-3)^2}\\\sf 4+(b-5)^2 & = \sf 25+(b-3)^2\\\sf 4+b^2-10b+25 & = \sf 25 + b^2-6b+9\\\sf b^2-10b+29 & = \sf b^2 -6b +34\\\sf -10b+29 & = \sf -6b + 34\\\sf -4b & = \sf 5\\\sf b & = -\dfrac{5}{4}\end{aligned}$