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ankoles [38]
3 years ago
15

A mailbox that is 4 1/2 feet tall casts a shadow that is 6 feet long. At the same time, a nearby ferris wheel casts a shadow 84

feet long. Find the height of the ferris wheel.
60 feet
Mathematics
2 answers:
gtnhenbr [62]3 years ago
8 0

Answer:

63 feet

Step-by-step explanation:

Trust me got it right

ira [324]3 years ago
4 0

Answer: 63 ft

Step-by-step explanation:

We can think of the shadow and height of objects as the catheti of a triangle rectangle.

We also know the relationship:

Tan(θ) = (opposite cathetus)/(adjacent cathetus)

if we define:

Tan(θ) = k (a real number, such that will be the same number for every object, because it depends on the angle θ that will depend on the position of the sun)

opposite cathetis = heigt.

adjacent cathetus = length of the shadow.

Then we have:

k = height/length of the shadow.

For the mailbox, we know that the height is (4 + 1/2) ft and the shadow is 6ft long, then we have the equation:

k = (4 + 1/2)ft/6ft = 0.75

Then for the ferris wheel we know:

shadow length = 84ft

we can write the same equation as above:

k = 0.75 = height/84ft

0.75*84ft = height = 63ft

The ferris wheel is 63 ft high.

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<h2>2 pizzas </h2>

Step-by-step explanation:

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3 0
3 years ago
Solve for x. will give brailiest<br><br> 2x + 5 = 27<br><br> 8.5<br><br> 11<br><br> 16<br><br> 64
Rama09 [41]
<span>2x + 5 = 27
subtract 5 to both sides
2x + 5 - 5 = 27 - 5
simplify
2x = 22
divide both sides by 2
2x/2 = 22/2
simplify
x = 11

answer is </span><span>11 (second choice)
</span>
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4 0
2 years ago
40 POINTS
dolphi86 [110]
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6 0
3 years ago
Read 2 more answers
30 points !!!!!!!!Solve the following problem. Show ALL steps to receive full credit.
Aleksandr [31]
7x - 8 = 8 + 3x
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5 0
3 years ago
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Let ρ = x3 + xe−x for x ∈ (0, 1), compute the center of mass.
hram777 [196]

The center of mass is mathematically given as

\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \  \ \  x \in(0,1)$ \\\end

We calculate the total mass of the system.

\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}

Step 03: Calculate the moment of the system.

\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}

we calculate the center of mass.

\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}

Read more about the center of mass.

brainly.com/question/27549055

#SPJ1

8 0
2 years ago
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