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3 years ago

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A monkey is swinging from the tree. On the first swing she passes through an arc of 20 m. With each swing she passes through an

arc 4/5 the length of the previous swing. How long was the monkeys 1st swing? What is the total distance the monkey has traveled when she completes her 10th swing?

2 answers:

anygoal [31]3 years ago

4 0

Answer:

Step-by-step explanation:

The answer is 89.26 m

Tema [17]3 years ago

4 0

Answer:

Step-by-step explanation:

=89

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The spread of a virus is modeled by V (t) = −t 3 + t 2 + 12t,

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Functions can be used to model real life scenarios

The reasonable domain is $\mathbf{[0,\infty)}$ .
The average rate of change from t = 0 to 2 is 20 persons per week
The instantaneous rate of change is $\mathbf{V'(t) = -3t^2 + 2t + 12}$ .
The slope of the tangent line at point (2,V(20) is 10
The rate of infection at the maximum point is 8.79 people per week

The function is given as:

$\mathbf{V(t) = -t^3 + t^2 + 12t}$

<u>(a) Sketch V(t)</u>

See attachment for the graph of $\mathbf{V(t) = -t^3 + t^2 + 12t}$

<u />

<u>(b) The reasonable domain</u>

t represents the number of weeks.

This means that: <em>t cannot be negative.</em>

So, the reasonable domain is: $\mathbf{[0,\infty)}$

<u />

<u>(c) Average rate of change from t = 0 to 2</u>

This is calculated as:

$\mathbf{m = \frac{V(a) - V(b)}{a - b}}$

So, we have:

$\mathbf{m = \frac{V(2) - V(0)}{2 - 0}}$

$\mathbf{m = \frac{V(2) - V(0)}{2}}$

Calculate <em>V(2) and V(0)</em>

$\mathbf{V(2) = (-2)^3 + (2)^2 + 12 \times 2 = 20}$

$\mathbf{V(0) = (0)^3 + (0)^2 + 12 \times 0 = 0}$

So, we have:

$\mathbf{m = \frac{20 - 0}{2}}$

$\mathbf{m = \frac{20}{2}}$

$\mathbf{m = 10}$

Hence, the average rate of change from t = 0 to 2 is 20

<u>(d) The instantaneous rate of change using limits</u>

$\mathbf{V(t) = -t^3 + t^2 + 12t}$

The instantaneous rate of change is calculated as:

$\mathbf{V'(t) = \lim_{h \to \infty} \frac{V(t + h) - V(t)}{h}}$

So, we have:

$\mathbf{V(t + h) = (-(t + h))^3 + (t + h)^2 + 12(t + h)}$

$\mathbf{V(t + h) = (-t - h)^3 + (t + h)^2 + 12(t + h)}$

Expand

$\mathbf{V(t + h) = (-t)^3 +3(-t)^2(-h) +3(-t)(-h)^2 + (-h)^3 + t^2 + 2th+ h^2 + 12t + 12h}$ $\mathbf{V(t + h) = -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h}$

Subtract V(t) from both sides

$\mathbf{V(t + h) - V(t)= -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h - V(t)}$

Substitute $\mathbf{V(t) = -t^3 + t^2 + 12t}$

$\mathbf{V(t + h) - V(t)= -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h +t^3 - t^2 - 12t}$

Cancel out common terms

$\mathbf{V(t + h) - V(t)= -3t^2h -3th^2 - h^3 + 2th+ h^2 + 12h}$

$\mathbf{V'(t) = \lim_{h \to \infty} \frac{V(t + h) - V(t)}{h}}$ becomes

$\mathbf{V'(t) = \lim_{h \to \infty} \frac{ -3t^2h -3th^2 - h^3 + 2th+ h^2 + 12h}{h}}$

$\mathbf{V'(t) = \lim_{h \to \infty} -3t^2 -3th - h^2 + 2t+ h + 12}$

Limit h to 0

$\mathbf{V'(t) = -3t^2 -3t\times 0 - 0^2 + 2t+ 0 + 12}$

$\mathbf{V'(t) = -3t^2 + 2t + 12}$

<u>(e) V(2) and V'(2)</u>

Substitute 2 for t in V(t) and V'(t)

So, we have:

$\mathbf{V(2) = (-2)^3 + (2)^2 + 12 \times 2 = 20}$

$\mathbf{V'(2) = -3 \times 2^2 + 2 \times 2 + 12 = 4}$

<em>Interpretation</em>

V(2) means that, 20 people were infected after 2 weeks of the virus spread

V'(2) means that, the rate of infection of the virus after 2 weeks is 4 people per week

<u>(f) Sketch the tangent line at (2,V(2))</u>

See attachment for the tangent line

The slope of this line is:

$\mathbf{m = \frac{V(2)}{2}}$

$\mathbf{m = \frac{20}{2}}$

$\mathbf{m = 10}$

The slope of the tangent line is 10

<u>(g) Estimate V(2.1)</u>

The <em>value of 2.1 </em>is

$\mathbf{V(2.1) = (-2.1)^3 + (2.1)^2 + 12 \times 2.1}$

$\mathbf{V(2.1) = 20.35}$

<u />

<u>(h) The maximum number of people infected at the same time</u>

Using the graph, the maximum point on the graph is:

$\mathbf{(t,V(t) = (2.361,20.745)}$

This means that:

The maximum number of people infected at the same time is approximately 21.

The rate of infection at this point is:

$\mathbf{m = \frac{V(t)}{t}}$

$\mathbf{m = \frac{20.745}{2.361}}$

$\mathbf{m = 8.79}$

The rate of infection is <em>8.79 people per week</em>

Read more about graphs and functions at:

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