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Dima020 [189]
3 years ago
9

A meter stick casts a shadow 2.4 m long at the same time a flagpole casts a shadow 9.7 m long. How tall is the flagpole

Mathematics
1 answer:
Luda [366]3 years ago
8 0
The first thing we are going to do is draw a diagram of the situation to help us to solve the situation.
We an draw tow similar triangle between the length of the shadows and the height of the stick and the pole.
Since <span>meter stick has a length of 1 meter, the height of our smaller triangle is 1 meter. Let </span>h be height of of the pole. Since both triangles are similiar we can establish a proportion between their corresponding sides and solve for h:
\frac{h}{1} = \frac{9.7}{2.4}
h=\frac{9.7}{2.4}
h=4.04

We can conclude that the pole is 4.04 meters tall. 

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Answer:

\large\boxed{(x-2)^2+(y-1)^2=34}

Step-by-step explanation:

The equation of a circle in standard form:

(x-h)^2+(y-k)^2=r^2

(h, k) - center

r - radius

We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)

Substitute:

h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

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

Substitute the coordinates of the points (2, 1) and (5, -4):

r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}

Finally we have:

(x-2)^2+(y-1)^2=(\sqrt{34})^2

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