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VLD [36.1K]
1 year ago
6

A stick 2 m long is placed vertically at point B. The top of the stick is in line with the top of a tree as seen from point A, w

hich is 3 m from the stick and 30 m from the tree. How tall is the tree?

Mathematics
1 answer:
Ber [7]1 year ago
6 0

Answer: 20 meters.

Step-by-step explanation:

1. Keeping on mind the information shown in the figure attached and the similarity of both triangles, you can calculate the height of the tree (h) as you can see below:

\frac{3}{30}=\frac{2}{h}

2. Now, you must solve for the height. So, you obtain the following result:

h=\frac{30*2}{3}\\h=20

3. Therefore, the height of the tree is 20 meters.

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She would be able to download 49 at this rate.
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Which equation combines with the given equation to form a system of equations with the solution x = 3 and y = 9? x + 2y = 21
max2010maxim [7]
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2 years ago
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Find the average rate of change of the function defined by the following table. x y -2 -5 2 35 6 75 10 115
Pani-rosa [81]

By definition we have that the average rate of change of the function is:

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 Evaluating the function for the complete interval we have that the AVR is given by:

 AVR = \frac{115 - (-5)}{10 - (-2)}

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6 0
1 year ago
The graph shows the solution to the initial value problem y'(t)=mt, y(t0)= -4
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y'(t)=mt
\\
\\ y=\int {mt} \, dt=m \int{t} \,dt=m \frac{t^2}{2} +C \\ \\y= \frac{mt^2}{2}+C

Now find equation of the graph. It passes through the point (2,3), and intersects y-axis at -2.

y=at^2+b
\\
\\3=a\times 2^2-2
\\
\\3=4a-2
\\
\\5=4a
\\
\\a= \frac{5}{4} 
\\
\\y=\frac{5}{4} t^2-2 \\ \\y= \frac{m}{2} t^2+C
\\ \\  \frac{m}{2} = \frac{5}{4} 
\\
\\m= \frac{5}{2} 
\\
\\C=-2
\\
\\y(t_0)=-4
\\
\\-4= \frac{5}{4} t_0^2-2
\\
\\ -2= \frac{5}{4} t_0^2
\\
\\ t_0^2=- \frac{8}{5} 
\\
\\t_0=\pm \sqrt{- \frac{8}{5} }



3 0
2 years ago
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