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4 years ago

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The Cool Company determines that the supply function for its basic air conditioning unit is S(p) = 10 + 0.002p3 and that its dem

and function is D(p) = 50 - 0.04p2, where p is the price. Determine the price for which the supply equals the demand.

2 answers:

Scorpion4ik [409]4 years ago

5 0

$21.86 is the answer that your looking for<span>
</span>

Radda [10]4 years ago

3 0

Answer: At $21.85, the supply will equal to demand.

Step-by-step explanation:

Since we have given that

Demand function is given by

$D(p) = 50 - 0.04p^2, \text{where p is the price.}$

Supply function is given by

$S(p) = 10 + 0.002p^3$

According to question, we need to find the price for which the supply equals the demand, i.e. Equilibrium price and quantity.

$D(p)=S(p)\\\\50-0.04p^2=10+0.002p^3\\\\50-10=0.002p^3+0.04p^2\\\\40=\frac{2}{1000}p^3+\frac{4}{100}p^2\\\\40=\frac{2}{100}p^2(\frac{1}{10}p+2)\\\\\frac{40\times 100}{2}=p^2(\frac{1p+20}{2})\\\\\mathrm{The\:Newton-Raphson\:method\:uses\:an\:iterative\:process\:to\:approach\:one\:root\:of\:a\:function}\\\\p\approx \:21.85861\dots$

So, at $21.85, the supply will equal to demand.

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Natasha_Volkova [10]

Answer:

Liam's rate = 3 meters per second

Edgar's rate = 3.2 meters per second

Edgar swims faster

Step-by-step explanation:

<u>Liam:</u>

The equation $d=3t$ gives the distance, d, in meters that Liam swims with respect to time, t, in seconds.

When $t=0,\ d=0.$

When $t=1,\ d=3.$

Rate of change:

$\dfrac{3-0}{1-0}=3$

Liam's rate is 3 metres per second.

<u>Edgar:</u>

When $t=20, \ d=64.$

When $t=40,\ t=128.$

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$\dfrac{128-64}{40-20}=\dfrac{64}{20}=3.2$

Edgar's rate is 3.2 metres per second.

Edgar swims faster.

5 0

3 years ago

A light bulb is designed by revolving the graph of:

nadya68 [22]

Answer:

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

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Algebra I</u>

Terms/Coefficients
Expand by FOIL (First Outside Inside Last)
Factoring

<u>Calculus</u>

Differentiation

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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>

Basic Power Rule: $\displaystyle y' = \frac{1}{2} \cdot \frac{1}{3}x^{\frac{1}{2} - 1} - \frac{3}{2} \cdot x^{\frac{3}{2} - 1}$
[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: $\displaystyle SA = 2\pi \int\limits^{\frac{1}{3}}_0 {(\frac{1}{3}x^{\frac{1}{2}} - x^{\frac{3}{2}}) \sqrt{\frac{(9x + 1)^2}{36x}} \, dx$
[Integral - Simplify]: $\displaystyle SA = 2\pi \int\limits^{\frac{1}{3}}_0 {-\frac{|9x + 1|(3x - 1)}{18}} \, dx$
[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>

[Integral] Define: $\displaystyle \int {|9x + 1|(3x - 1)} \, dx$
[Integral] Assumption of Positive/Correction Factors: $\displaystyle \frac{9x + 1}{|9x + 1|} \int {(9x + 1)(3x - 1)} \, dx$
[Integral] Expand - FOIL: $\displaystyle \frac{9x + 1}{|9x + 1|} \int {27x^2 - 6x - 1} \, dx$
[Integral] Integrate - Basic Power Rule: $\displaystyle \frac{9x + 1}{|9x + 1|} (9x^3 - 3x^2 - x)$
[Expression] Multiply: $\displaystyle \frac{(9x + 1)(9x^3 - 3x^2 - x)}{|9x + 1|}$

<u>Step 5: Integrate Pt. 3</u>

[Integral] Substitute/Integral - FTC: $\displaystyle SA = \frac{- \pi}{9} (\frac{(9x + 1)(9x^3 - 3x^2 - x)}{|9x + 1|})|\limits_{0}^{\frac{1}{3}}$
[Integrate] Evaluate FTC: $\displaystyle SA = \frac{- \pi}{9} (\frac{-1}{3})$
[Expression] Multiply: $\displaystyle SA = \frac{\pi}{27} \ ft^2$

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

<u>Step 6: Find Amount of Glass</u>

<em>Convert ft² to in² and multiply by 0.015 in (given) to find amount of glass.</em>

Convert ft² to in²: $\displaystyle \frac{\pi}{27} \ ft^2 \ \div 144 \ in^2/ft^2 = \frac{16 \pi}{3} \ in^2$
Multiply: $\displaystyle \frac{16 \pi}{3} \ in^2 \cdot 0.015 \ in = 0.251327 \ in. \ of \ glass$

And we have our final answer! Hope this helped on your Calc BC journey!

5 0

3 years ago

The long jump distances of four students are listed in the table below. How much farther

Fiesta28 [93]

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6-4=2

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3 years ago

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Alexandra [31]

Step-by-step explanation:

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