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const2013 [10]
4 years ago
11

What is the approximate circumference of a circle with a radius of 40 inches? Use π ≈ 3.14.

Mathematics
1 answer:
Vlada [557]4 years ago
6 0
To find cirfumfrence we use the formula C = 2 \pi r
Soooo
\pi = 3.14
r = 40 inches
So now we plug it in :P
C = 2 x 3.14 x 40 = 251.2 
answer = B.) 251.2 inches 
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I need someone to do this question
vfiekz [6]

9514 1404 393

Answer:

  $1487.50

Step-by-step explanation:

The amount of interest due is ...

  I = Prt

where P is the loan amount, r is the annual rate, and t is the number of years. Here, t = 6 months = 1/2 year, so the interest due is ...

  I = $1400×0.125×1/2 = $87.50

The total amount due is the sum of the loan amount and the interest:

  due = $1400 +87.50 = $1487.50

The total amount due after 60 months is $1487.50.

8 0
3 years ago
Read 2 more answers
Hi, how do we do this question?​
Nutka1998 [239]

Answer:

\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C

General Formulas and Concepts:

<u>Algebra I</u>

  • Terms/Coefficients
  • Factoring

<u>Algebra II</u>

  • Polynomial Long Division

<u>Calculus</u>

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:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals
  • Integration Constant C
  • Indefinite Integrals

Integration Rule [Reverse Power Rule]:                                                               \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Property [Multiplied Constant]:                                                         \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

Integration Property [Addition/Subtraction]:                                                       \displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx

Logarithmic Integration

U-Substitution

Step-by-step explanation:

*Note:

You could use u-solve instead of rewriting the integrand to integrate this integral.

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \int {\frac{2x}{3x + 1}} \, dx

<u>Step 2: Integrate Pt. 1</u>

  1. [Integrand] Rewrite [Polynomial Long Division (See Attachment)]:           \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\bigg( \frac{2}{3} - \frac{2}{3(3x + 1)} \bigg)} \, dx
  2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\frac{2}{3}} \, dx - \int {\frac{2}{3(3x + 1)}} \, dx
  3. [Integrals] Rewrite [Integration Property - Multiplied Constant]:               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}\int {} \, dx - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx
  4. [1st Integral] Reverse Power Rule:                                                               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx

<u>Step 3: Integrate Pt. 2</u>

<em>Identify variables for u-substitution.</em>

  1. Set <em>u</em>:                                                                                                             \displaystyle u = 3x + 1
  2. [<em>u</em>] Differentiate [Basic Power Rule]:                                                             \displaystyle du = 3 \ dx

<u>Step 4: Integrate Pt. 3</u>

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{3}{3x + 1}} \, dx
  2. [Integral] U-Substitution:                                                                               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{1}{u}} \, du
  3. [Integral] Logarithmic Integration:                                                               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|u| + C
  4. Back-Substitute:                                                                                            \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|3x + 1| + C
  5. Factor:                                                                                                           \displaystyle \int {\frac{2x}{3x + 1}} \, dx = -2 \bigg( \frac{1}{9}ln|3x + 1| - \frac{x}{3}  \bigg) + C
  6. Rewrite:                                                                                                         \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C

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

Unit: Integration

Book: College Calculus 10e

8 0
3 years ago
Find the quotient of 2.9x10 and 4x10.-9​
Natalka [10]

Answer:

??

2.9×10 = 29

4×10.-9 =31

8 0
3 years ago
Image point B (4,-8) was transformed using the translation (x-2,y+3) what were the coordinates of B?
PilotLPTM [1.2K]
The coordinates of the image point are given to be B(4, -8). We are to find the coordinates of pre-image that is the coordinates before the translation.

The translation made was (x-2, y+3) 

This means, the x coordinate was moved 2 units to left and y coordinate was moved 3 units above. 

So, the coordinates of original point will be, 2 units to right of B and 3 units down of B.

So the coordinates will be (6, -11)
 
5 0
3 years ago
Read 2 more answers
adriano runs a website that helps writers write better stories. every month, adriano recieves a subscription fee of $5 from each
ZanzabumX [31]
200 + 40 - 10 = 230
230 subs * $5 = $1150 earned
6 0
4 years ago
Read 2 more answers
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