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3 years ago

I need help with 2 plz help

Mathematics

1 answer:

Tatiana [17]3 years ago

7 0

Answer:

a=31.61

Step-by-step explanation:

Pythagorean theorem states that square hypotenuse side is equal to square opposite side plus square adjacent side

Hyp²= Opp²+Adj²

the opposite is represented by a in the question, so

60²=a²+51²

3600=a²+2601

combine the like terms

3600-2601=a²

999=a²

a²=999

square root both sides to remove the square

√a²=√999 {square cancels the square root

a=√999

a=31.606

a=31.61 approximately

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Which relation is a function? (look at image below)

Dimas [21]

Answer:

Bottom Right Graph (Absolute Value)

General Formulas and Concepts:

Vertical Line Test

Step-by-step explanation:

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2 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:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

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

Calculus

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:

Step 1: Define

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

Step 2: Differentiate

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: $\displaystyle y' = \frac{1}{6\sqrt{x}} - \frac{3\sqrt{x}}{2}}$

Step 3: Integrate Pt. 1

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: $\displaystyle SA = \frac{- \pi}{9} \int\limits^{\frac{1}{3}}_0 {|9x + 1|(3x - 1)} \, dx$

Step 4: Integrate Pt. 2

[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: $\displaystyle \frac{(9x + 1)(9x^3 - 3x^2 - x)}{|9x + 1|}$

Step 5: Integrate Pt. 3

[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: $\displaystyle SA = \frac{\pi}{27} \ ft^2$

It is in ft² because it is given that our axis are in ft.

Step 6: Find Amount of Glass

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

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