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2 years ago

What number is equivalent to 43/12

Mathematics

2 answers:

jok3333 [9.3K]2 years ago

8 0

43/12 can be
3 7/12

Hope this helps

Stells [14]2 years ago

8 0

43/12 as a mixed number is 3 7/12

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2 years ago

A large spherical structure has a 37.5-foot radius. The sphere is covered with panels. If each panel has an area of about 49 ft2

Citrus2011 [14]

Answer:

Option (1)

Step-by-step explanation:

Radius of the spherical structure = 37.5 feet

Surface area of each panel = 49 square feet

Sphere is covered with panels.

Therefore, area of the panels needed = $\frac{\text{Total surface area of the sphere}}{\text{Area of one panel}}$

Surface area of the sphere = 4πr²

Here 'r' = radius of the sphere

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Surface area of the sphere = 4π(37.5)²

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Number of panels required to cover the sphere = $\frac{17671.5}{49}$

= 360

Therefore, Option (1) will be the answer.

4 0

2 years ago

Read 2 more answers

Help differentiate this

Arte-miy333 [17]

Answer:

$\displaystyle y' = 20x^3 + 6x^2 + 70x + 9$

General Formulas and Concepts:

Pre-Algebra

Distributive Property

Algebra I

Terms/Coefficients
Expand by FOIL
Functions
Function Notation

Calculus

Derivatives

Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = (x^3 + 7x - 1)(5x + 2)$

Step 2: Differentiate

Product Rule: $\displaystyle y' = \frac{d}{dx}[(x^3 + 7x - 1)](5x + 2) + (x^3 + 7x - 1)\frac{d}{dx}[(5x + 2)]$
Basic Power Rule [Derivative Property - Addition/Subtraction]: $\displaystyle y' = (3x^{3 - 1}+ 7x^{1 - 1} - 0)(5x + 2) + (x^3 + 7x - 1)(5x^{1 - 1} + 0)$
Simplify: $\displaystyle y' = (3x^2+ 7)(5x + 2) + 5(x^3 + 7x - 1)$
Expand: $\displaystyle y' = 15x^3 + 6x^2 + 35x + 14 + 5(x^3 + 7x - 1)$
[Distributive Property] Distribute 5: $\displaystyle y' = 15x^3 + 6x^2 + 35x + 14 + 5x^3 + 35x - 5$
Combine like terms: $\displaystyle y' = 20x^3 + 6x^2 + 70x + 9$