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Artist 52 [7]
3 years ago
7

24 divide by blank plus 5 multipled by 3 plus 2 equals 20

Mathematics
1 answer:
sineoko [7]3 years ago
7 0

Answer:

8

Step-by-step explanation:

The answer would be 8 because the question would be

24 ÷ x + 5 × 3 + 2=20  right?

The first thing you would want to do is multiply the 5 and the 3 because of PEMDAS which should give you  24 ÷ x + 15 + 2 = 20.

15 plus 2 is 17 and then you would subtract 17 from 20 which would give you 3. So you know know that 24÷x has to be 3 and 24÷3 is 8 so x=8.

Hope this helped!

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Step-by-step explanation:

2.5 = 5 * .5

1 = 1

70 = 2 * 5 *  7

LCM = 2 * 5 * 7

If you include the 1/2, you will reduce the LCM to 35, but 70 will be left out of the LCM.

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<span>in order least to greatest
</span>-5/6, <span>-1/16, 5/3

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What is the length of the curve with parametric equations x = t - cos(t), y = 1 - sin(t) from t = 0 to t = π? (5 points)
zzz [600]

Answer:

B) 4√2

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Parametric Differentiation

Integration

  • Integrals
  • Definite Integrals
  • Integration Constant C

Arc Length Formula [Parametric]:                                                                         \displaystyle AL = \int\limits^b_a {\sqrt{[x'(t)]^2 + [y(t)]^2}} \, dx

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \left \{ {{x = t - cos(t)} \atop {y = 1 - sin(t)}} \right.

Interval [0, π]

<u>Step 2: Find Arc Length</u>

  1. [Parametrics] Differentiate [Basic Power Rule, Trig Differentiation]:         \displaystyle \left \{ {{x' = 1 + sin(t)} \atop {y' = -cos(t)}} \right.
  2. Substitute in variables [Arc Length Formula - Parametric]:                       \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{[1 + sin(t)]^2 + [-cos(t)]^2}} \, dx
  3. [Integrand] Simplify:                                                                                       \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx
  4. [Integral] Evaluate:                                                                                         \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}

Topic: AP Calculus BC (Calculus I + II)

Unit: Parametric Integration

Book: College Calculus 10e

4 0
2 years ago
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