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elixir [45]
3 years ago
13

Two side lengths of a regular polygon are represented by the expressions 7x+8 and 13x-4

Mathematics
1 answer:
shtirl [24]3 years ago
5 0

Answer:hi

Step-by-step explanation:

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Please answer and explain the link below
Leviafan [203]

Answer:

See explanation.

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • Functions
  • Function Notation

<u>Calculus</u>

Limits

  • Right-Side Limit:                                                                                             \displaystyle  \lim_{x \to c^+} f(x)
  • Left-Side Limit:                                                                                               \displaystyle  \lim_{x \to c^-} f(x)

Limit Rule [Variable Direct Substitution]:                                                             \displaystyle \lim_{x \to c} x = c

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle f(x) = \left \{ {{\sqrt{x + 1}, \ x < 3} \atop {5 - x, \ x \geq 3}} \right.

<u>Step 2: Find Right Limit</u>

  1. Substitute in variables [Right-Side Limit]:                                                      \displaystyle  \lim_{x \to 3^+} 5 - x
  2. Evaluate limit [Limit Rule - Variable Direct Substitution]:                           \displaystyle  \lim_{x \to 3^+} 5 - x = 5 - 3
  3. Subtract:                                                                                                         \displaystyle  \lim_{x \to 3^+} 5 - x = 2

∴ the right-side limit equals 2.

<u>Step 3: Find Left Limit</u>

  1. Substitute in variables [Left-Side Limit]:                                                       \displaystyle  \lim_{x \to 3^-} \sqrt{x + 1}
  2. Evaluate limit [Limit Rule - Variable Direct Substitution]:                           \displaystyle  \lim_{x \to 3^-} \sqrt{x + 1} = \sqrt{3 + 1}
  3. [√Radical] Add:                                                                                             \displaystyle  \lim_{x \to 3^-} \sqrt{x + 1} = \sqrt{4}
  4. [√Radical] Evaluate:                                                                                       \displaystyle  \lim_{x \to 3^-} \sqrt{x + 1} = 2

∴ the left-side limit equals 2.

<u>Step 4: Find Limit</u>

<em>The right and left-side limits are equal.</em>

∴  \displaystyle  \lim_{x \to 3} f(x) = 2

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

Unit: Limits

Book: College Calculus 10e

4 0
3 years ago
Rebecca sells greetings cards at £1.47 each. She donates 3 7 of this money to a charity. How much does the charity receive from
ANTONII [103]
The charity receive £54.39 form the sale of each card.
4 0
3 years ago
An indoor running track is 200 meters in length. During a 3,000-meter race, runners must complete 15 laps of the track. An elect
BartSMP [9]

Answer:

D

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
If we approximate the function y=sin(x) with a0+a1 x a2 x^2 +a3 x^3, what is a0,a2,a2,a3?
Rashid [163]

The coefficients a_0,a_1,a_2,a_3 could be chosen to be the coefficients in the Maclaurin series of \sin(x).

We have

y = \sin(x) \approx a_0 + a_1 x + a_2 x^2 + a_3 x^3 \\\\ \implies y(0) = 0 = a_0

y' = \cos(x) \approx a_1 + 2a_2 x + 3a_3 x^2 \\\\ \implies y'(0) = 1 = a_1

y'' = -\sin(x) \approx 2a_2 + 6a_3 x \\\\ \implies y''(0) = 0 = 2a_2

y''' = -\cos(x) \approx 6a_3 \\\\ \implies y'''(0) = -1 = 6a_3

It follows that a_0=0, a_1=1, a_2=0, and a_3 = -\frac16.

7 0
2 years ago
4
Alex Ar [27]

Answer:

jh

Step-by-step explanation:

hh6bynnynyb7j7jbtvrf2xecrvubuvrcryvhbjbibxidh ixh idh!7sgzuefzuwz

5 0
3 years ago
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