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geniusboy [140]

3 years ago

10

A 20 foot ladder leaning against a wall is used to reach a window that is 17 feet above the ground. How far from the wall is the

bottom of the ladder? Round to the nearest tenth of a foot.

1 answer:

GarryVolchara [31]3 years ago

8 0

The bottom of the ladder is 10.5 feet away from the wall

Step-by-step explanation:

The given scenario forms a right triangle.

Where

The length of ladder will be the hypotenuse

The wall on which the window is situated will ebt he perpendicular and

The distance between the foot of ladder and the wall will be the base

So,

Hypotenuse = H = 20 foot

Perpendicular = P = 17 feet

Base = B = ?

Using the Pythagoras theorem

$H^2=P^2+B^2\\(20)^2=(17)^2+B^2\\400=289+B^2\\400-289=B^2\\B^2=111\\Taking\ square\ root\ on\ both\ sides\\\sqrt{B^2}=\sqrt{111}\\B=10.53\\Rounding\ off\ to\ the\ nearest\ tenth\\B=10.5$

The bottom of the ladder is 10.5 feet away from the wall

Keywords: Triangle, Pythagoras Theorem

Learn more about Pythagoras theorem at:

#LearnwithBrainly

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A light bulb is designed by revolving the graph of:

nadya68 [22]

Answer:

$\displaystyle 0.251327 \ in. \ of \ glass$

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<u>Calculus</u>

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Derivative Notation

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Integration

Integration Property: $\displaystyle \int\limits^a_b {cf(x)} \, dx = c \int\limits^a_b {f(x)} \, dx$

Fundamental Theorem of Calculus: $\displaystyle \int\limits^a_b {f(x)} \, dx = F(b) - F(a)$

Area between Two Curves

Volumes of Revolution

Arc Length Formula: $\displaystyle AL = \int\limits^a_b {\sqrt{1+ [f'(x)]^2}} \, dx$

Surface Area Formula: $\displaystyle SA = 2\pi \int\limits^a_b {f(x) \sqrt{1+ [f'(x)]^2}} \, dx$

Step-by-step explanation:

<u>Step 1: Define</u>

$\displaystyle y = \frac{1}{3}x^{\frac{1}{2}} - x^{\frac{3}{2}}\\Interval: [0, \frac{1}{3}]$

<u>Step 2: Differentiate</u>

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[Derivative] Simplify: $\displaystyle y' = \frac{1}{6}x^{\frac{-1}{2}} - \frac{3}{2}x^{\frac{1}{2}}$
[Derivative] Simplify: $\displaystyle y' = \frac{1}{6\sqrt{x}} - \frac{3\sqrt{x}}{2}}$

<u>Step 3: Integrate Pt. 1</u>

Substitute [Surface Area]: $\displaystyle SA = 2\pi \int\limits^{\frac{1}{3}}_0 {(\frac{1}{3}x^{\frac{1}{2}} - x^{\frac{3}{2}}) \sqrt{1+ [\frac{1}{6\sqrt{x}} - \frac{3\sqrt{x}}{2}}]^2}} \, dx$
[Integral - √Radical] Expand/Add: $\displaystyle SA = 2\pi \int\limits^{\frac{1}{3}}_0 {(\frac{1}{3}x^{\frac{1}{2}} - x^{\frac{3}{2}}) \sqrt{\frac{81x^2+18x+1}{36x}} \, dx$
[Integral - √Radical] Factor: $\displaystyle SA = 2\pi \int\limits^{\frac{1}{3}}_0 {(\frac{1}{3}x^{\frac{1}{2}} - x^{\frac{3}{2}}) \sqrt{\frac{(9x + 1)^2}{36x}} \, dx$
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[Integral] Integration Property: $\displaystyle SA = \frac{- \pi}{9} \int\limits^{\frac{1}{3}}_0 {|9x + 1|(3x - 1)} \, dx$

<u>Step 4: Integrate Pt. 2</u>

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<em>It is in ft² because it is given that our axis are in ft.</em>

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Convert ft² to in²: $\displaystyle \frac{\pi}{27} \ ft^2 \ \div 144 \ in^2/ft^2 = \frac{16 \pi}{3} \ in^2$
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And we have our final answer! Hope this helped on your Calc BC journey!

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3 years ago

Consider the equation 5+ x=n what must be true about any value of x if n is a negative number

nlexa [21]

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malfutka [58]

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6 0

3 years ago

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nevsk [136]

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Read 2 more answers