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Elden [556K]
3 years ago
7

Aaden is 1.75 meters tall. At 11 a.m., he measures the length of a tree's shadow to be 37.65 meters. He stands 32.9 meters away

from the tree, so that the tip of his shadow meets the tip of the tree's shadow. Find the height of the tree to the nearest hundredth of a meter.
Mathematics
1 answer:
Ratling [72]3 years ago
8 0

The relationship between Aaden and the tree's height is an illustration of equivalent ratio

The height of the tree is 2.00 meters

At 11 a.m, we have:

\mathbf{Aaden = 1.75m}

\mathbf{Tree\ Shadow = 37.65m}

\mathbf{Aaden\ Shadow = 32.9m}

So, we make use of the following equivalent ratio

\mathbf{Aaden : Tree = Aaden\ Shadow : Tree\ Shadow}

This gives

\mathbf{1.75: Tree = 32.9: 37.65}

Express as fractions

\mathbf{\frac{Tree}{1.75} = \frac{37.65}{32.9}}

Multiply both sides by 1.75

\mathbf{Tree = \frac{37.65}{32.9} \times 1.75}

\mathbf{Tree = 2.00}

Hence, the height of the tree is 2.00 meters

Read more about equivalent ratios at:

brainly.com/question/18441891

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Alex Ar [27]
When we approach limits, we are finding values that are infinitesimally approaching this x-value. Essentially, we consider the approximate location that this root or limit appears. This is essential when it comes to taking Calculus, and finding the limit or rate of change of a function.

When we are attempting limits questions, there are several tests we attempt first.

1. Evaluate the limit by substituting the value of the x-value as it approaches the value (direct evaluation of a limit)
2. Rearrangement of the function, such that we can evaluate the limit.
3. (TRIGONOMETRIC PROPERTIES)
\lim_{x \to 0} (\frac{sinx}{x}) = 1
\lim_{x \to 0} (\frac{tanx}{x}) = 1
4. Using L'Hopital's Rule for indeterminate limits, such as 0/0, -infinity/infinity, or infinity/infinity.

For example:

1) \lim_{x \to 0}\frac{\sqrt{x} - 5}{x - 25}

We can do this using the first and second method.
<em>Method 1: Direct evaluation:</em>

Substitute x = 0 to the function.
\frac{\sqrt{0} - 5}{0 - 25}
= \frac{-5}{-25}
= \frac{1}{5}

<em>Method 2: Rearranging the function
</em>

We can see that x - 25 can be rewritten as: (√x - 5)(√x + 5)
By rewriting it in this form, the top will cancel with the bottom easily, and our limit comes out the same.

\lim_{x \to 0}\frac{(\sqrt{x} - 5)}{(\sqrt{x} - 5)(\sqrt{x} + 5)}
= \lim_{x \to 0}\frac{1}{(\sqrt{x} + 5)}}
= \frac{1}{5}

Every example works exactly the same way, and by remembering these criteria, every limit question should come out pretty naturally.
8 0
3 years ago
What is the unit rate for meters per second if a car travels 450 meters in 25 ​seconds?
Fantom [35]
M/s =meters per second
7 0
3 years ago
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A lawn mower is pushed a distance of 100 ft. Along a horizontal path by a constant force of 60 lbs. Yhe handle of the lawn mower
stira [4]
For this case we have the following equation:
 w = F • PQ
 Where,
 w: work done
 F: is the force vector
 PQ: is the vector of the direction of movement.
 Rewriting the equation we have:
 w = || F || • || PQ || costheta
 Substituting values:
 w = (60) * (100) * (cos (45))
 w = (60) * (100) * (root (2) / 2)
 w = 4242.640687 lb.ft
 Answer:
 
The work done pushing the lawn mower is:
 
w = 4242.6 lb.ft
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3 years ago
Write an equation for this trend line
jeka94

Answer:

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Step-by-step explanation:

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Scorpion4ik [409]

Answer:g(X)=2x^2-10

Step-by-step explanation:

If a function f(x) is translated 'c' units down then the equation of the translated function be f(x)-c

Here, f(x) is translated 2 units down to obtain g(x).

Hence, we have

g(x)=f(x)-2

   =2x^2-8-2

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