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3 years ago

What is the circumference of earth's equator?

Mathematics

1 answer:

AlekseyPX3 years ago

3 0

Answer:

40,075 kilometers

Step-by-step explanation:

The distance around the Earth at the Equator, its circumference, is 40,075 kilometers

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I need help with math!!! time is running out!!! will mark brainliest!!!

fenix001 [56]

I see the question last and the 4 choices first.
Look at the points you have in the table with the question.
(0, 0), (4, 2), (9, 3)
The only graph that contains those three points is the second option.

4 0

3 years ago

What is the probability that 1 or 2 are rolled on a number cube with sides numbered 1, 2, 3, 4,

choli [55]

Answer: $\frac{1}{3}$

Step-by-step explanation:

total sides = 6

p (rolling a 1) = $\frac{1}{6}$

p (rolling a 2) = $\frac{1}{6\\}$

Note:

or - add

and - multiply

∴ $\frac{1}{6} +\frac{1}{6} = \frac{2}{6}$

∴ $\frac{2}{6} = \frac{1}{3}$

hence, p (rolling a 1 or 2) = $\frac{1}{3}$

4 0

2 years ago

What is the value of x?

Helen [10]

They should be the same since it’s congruent ??

6 0

3 years ago

4. What is the circumference of a circle whose area is 81π cm squared?

vichka [17]

Area = pi*r^2
Circumference = 2pi*r

Step 1: find the radius from area
81pi=pi*r^2
81=r^2
9=r

Step 2: plug in radius to the circumference equation

Circumference = 2pi*r

18pi is the answer, otherwise know as 56.52 as well.

7 0

2 years ago

Read 2 more answers

Hello again! This is another Calculus question to be explained.

podryga [215]

Answer:

See explanation.

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions

Function Notation

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

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

Step-by-step explanation:

We are given the following and are trying to find the second derivative at x = 2:

$\displaystyle f(2) = 2$

$\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}$

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

$\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}$

When we differentiate this, we must follow the Chain Rule: $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]$

Use the Basic Power Rule:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')$

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]$

Simplifying it, we have:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]$

We can rewrite the 2nd derivative using exponential rules:

$\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}$

To evaluate the 2nd derivative at x = 2, simply substitute in x = 2 and the value f(2) = 2 into it:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}$

When we evaluate this using order of operations, we should obtain our answer:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

5 0

2 years ago