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3 years ago

A motorcar is running at the rate of 66 km per hour. Each

Mathematics

1 answer:

amm18123 years ago

3 0

Answer:

Given:

The radius of a wheel =r=35 cm=0.35 cm.

The speed it must keep =s=66 km/h=

66×1000

m/min=1100 m/min

To find out:

The number of revolution =n, it makes per minute to maintain that speed.

Solution:

The circumference =C of the wheel =2πr=2×

×0.35 m=2.2 m=C

The distance the wheel covers in 1 min=d=1100 m.

Now the distance =d covered by a wheel in one revolution = the circumference of the wheel.

∴ Here the number of revolution =n=

2.2

1100

=500.

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3 years ago

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The ratio of the side lengths of a quadrilateral is 3:3:5:8, and the perimeter is 380cm. What is the measure of the longest side

castortr0y [4]

Answer:

160

Step-by-step explanation:

Add the ratios together to get the sum of them is 19. Since the perimeter is 380, divide 380 by 19 to get 20.

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3 years ago

What are two expressions that are equivalent to 6(5j + 2 + j)?

ziro4ka [17]

Answer:

1. 30j+12+6j

2. 36j+12

3. 6(6j+2)

Step-by-step explanation:

1. You just multiply so... 6x5j ... 6x2 ... 6xj

2. you multiply and then simplify so first you have 30j+12+6j, so you add like terms so... 30j+6j= 36j and then +12

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3 years ago

What is the name of segment BD ?!

Alenkasestr [34]

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Step-by-step explanation:

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3 years ago

Implicit differentiation Please help

Anvisha [2.4K]

Answer:

$y''(-1) =8$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Algebra I

Factoring

Calculus

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

-xy - 2y = -4

Rate of change of the tangent line at point (-1, 4)

Step 2: Differentiate Pt. 1

Find 1st Derivative

Implicit Differentiation [Product Rule/Basic Power Rule]: $-y - xy' - 2y' = 0$
[Algebra] Isolate y' terms: $-xy' - 2y' = y$
[Algebra] Factor y': $y'(-x - 2) = y$
[Algebra] Isolate y': $y' = \frac{y}{-x-2}$
[Algebra] Rewrite: $y' = \frac{-y}{x+2}$

Step 3: Find y

Define equation: $-xy - 2y = -4$
Factor y: $y(-x - 2) = -4$
Isolate y: $y = \frac{-4}{-x-2}$
Simplify: $y = \frac{4}{x+2}$

Step 4: Rewrite 1st Derivative

[Algebra] Substitute in y: $y' = \frac{-\frac{4}{x+2} }{x+2}$
[Algebra] Simplify: $y' = \frac{-4}{(x+2)^2}$

Step 5: Differentiate Pt. 2

Find 2nd Derivative

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{0(x+2)^2 - 8 \cdot 2(x + 2) \cdot 1}{[(x + 2)^2]^2}$
[Derivative] Simplify: $y'' = \frac{8}{(x+2)^3}$

Step 6: Find Slope at Given Point

[Algebra] Substitute in x: $y''(-1) = \frac{8}{(-1+2)^3}$
[Algebra] Evaluate: $y''(-1) =8$

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3 years ago

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