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2 years ago

Can someone help me with this question?

Mathematics

1 answer:

meriva2 years ago

7 0

Answer:

m = 4

Step-by-step explanation:

$\because ABCD\sim QRST$

$\therefore \frac{10}{m}= \frac{15}{6}$

$\therefore m= \frac{10\times 6}{15}$

$\therefore m= \frac{60}{15}$

$\therefore m=4$

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The equation of a circle is given below. Identify the center and radius. Then graph the circle.

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Answer:

Center = (3, 0)
Radius = 3
Graph = see below

Concept:

Here, we need to know the idea of the circle equation.

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A sailboat is sitting at rest near its dock. A rope attached to the bow of the boat is drawn in over a pulley that stands on a p

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Answer:

The boat is approaching the dock at rate of 2.14 ft/s

Step-by-step explanation:

The situation given in the question can be modeled as a triangle, please refer to the attached diagram.

A rope attached to the bow of the boat is drawn in over a pulley that stands on a post on the end of the dock that is 5 feet higher than the bow that means x = 5 ft.

The length of rope from bow to pulley is 13 feet that means y = 13 ft.

We know that Pythagorean theorem is given by

$x^{2} + y^{2} = z^{2}$

Differentiating the above equation with respect to time yields,

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}$

$x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$

dx/dt = 0 since dock height doesn't change

$y\frac{dy}{dt} = z\frac{dz}{dt}$

$\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}$

The rope is being pulled in at a rate of 2 feet per second that is dz/dt = 2 ft/s

First we need to find z

z² = (5)² + (13)²

z² = 194

z = √194

z = 13.93 ft

So,

$\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}$

$\frac{dy}{dt} = \frac{13.93}{13}(2)$

$\frac{dy}{dt} = 2.14$ $ft/s$

Therefore, the boat is approaching the dock at rate of 2.14 ft/s

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