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neonofarm [45]
3 years ago
6

Why using the order of operations is important.

Mathematics
1 answer:
Otrada [13]3 years ago
7 0

Answer:

Simply: It makes sure you get the correct answer

Step-by-step explanation:

Using order of operations (pemdas) ensures you do the correct operations in the correct order. If not, then everyone could get loads of different answers for the same question. It provides order in the math world.

Hope this helps!

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3/4 ×2/2 to get the denominator to 8

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I don’t understand this question.
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The answer is 106 degrees

Step-by-step explanation:

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IF ANYONE HELPS ME WITH THIS I WILL GIVE U BRAINLIEST AND 100 POINTS BTW ITS INTEGRALS FOR CALCULUS
dlinn [17]

Answer:

\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{e^\bigg{\frac{1}{4}}}{8} - \frac{e^\bigg{\frac{1}{9}}}{8}

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>

  • Terms/Coefficients
  • Factoring
  • Exponential Rule [Rewrite]:                                                                           \displaystyle b^{-m} = \frac{1}{b^m}

<u>Calculus</u>

Derivatives

Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integrals

  • Definite Integrals

Integration Constant C

Integration Rule [Fundamental Theorem of Calculus 1]:                                     \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

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

U-Substitution

eˣ Integration:                                                                                                         \displaystyle \int {e^u} \, dx = e^u + C

Step-by-step explanation:

<u>Step 1: Define</u>

\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx

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

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

  1. Set:                                                                                                                 \displaystyle u = 4x^{-2}
  2. [<em>u</em>] Differentiate [Derivative Rule - Basic Power Rule]:                               \displaystyle du = -8x^{-3} \ dx
  3. [<em>du</em>] Rewrite [Exponential Rule - Rewrite]:                                                   \displaystyle du = \frac{-8}{x^3} \ dx

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

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 \displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^6_4 {\frac{-8}{x^3}e^{4x^{-2}}} \, dx
  2. [Integral] U-Substitution:                                                                                \displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{1}{9}}_{\frac{1}{4}} {e^u} \, dx
  3. [Integral] eˣ Integration:                                                                                \displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8}(e^u) \bigg| \limits^{\frac{1}{9}}_{\frac{1}{4}}
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:          \displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8} \bigg[ -e^\bigg{\frac{1}{9}} \bigg( e^\bigg{\frac{5}{36}} - 1 \bigg) \bigg]
  5. Simplify:                                                                                                         \displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{e^\bigg{\frac{1}{4}}}{8} - \frac{e^\bigg{\frac{1}{9}}}{8}

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

Unit: Integration

Book: College Calculus 10e

5 0
2 years ago
A camera priced at ​$220 is marked down 30​%. Find the price of the camera after the markdown.
Usimov [2.4K]

Answer:

154

Step-by-step explanation:

220-(220*30%)=154

Use this on google next time so you do not waste points on brainly

6 0
3 years ago
Ian wants to promote his band on the internet. Site a offers website hosting for $4.95 per month with a $49.99 startup fee. Site
olya-2409 [2.1K]
 <span>I believe for twelve (12) months. 

Site A: $49.95 plus $59.40(4.95 x 12) equals $109.35 

Site B: $9.95 x 12 equals $119.40 

No, wait that's not right. 

Okay, at 8 months, site A is pretty much at $90 (forget the nickels) and site B is $80 so site B is less. 
At 7 months, site A is $85 and site B is $70 so site B is less. 
At 9 months, site A is $95 and site B is $90 so site B is less. 
At 10 months, site A $100 and site B is $100 

It's got to be around 10 months somewhere. 

Ten months would be $99.45 for site A and $99.50 for site B so B is less. 

Eleven months is $104.40 for A and $109.45 for B so now B is more.</span>
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