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ikadub [295]
2 years ago
13

Help with this question (100 points)

Mathematics
1 answer:
Masteriza [31]2 years ago
6 0

Answer:

y = \frac{\pi}{3}

Step-by-step explanation:

using the addition formula for sine

sin(x + y) = sinxcosy + cosxsiny

then

sinxcosy + cosxsiny = \frac{1}{2} sinx + \frac{\sqrt{3} }{2} cosx

for the 2 sides to be equal , then

cosy = \frac{1}{2} and siny = \frac{\sqrt{3} }{2}

then

y = cos^{-1} (\frac{1}{2} ) = \frac{\pi }{3}

and

y = sin^{-1} ( \frac{\sqrt{3} }{2} ) = \frac{\pi }{3}

thus y = \frac{\pi }{3}

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3 years ago
Given that f(–2.4) = -1 and f(-1.9) = -8, approximate<br> f'(-2.4).<br> f'(-2.4)
lora16 [44]

Answer:

f'(-2.4) ≈ -14

General Formulas and Concepts:
<u>Algebra I</u>

Coordinate Planes

  • Coordinates (x, y)

Slope Formula: \displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}

Functions

  • Function Notation

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Step-by-step explanation:

*Note:

The definition of a derivative is the slope of the <em>tangent</em> <em>line</em>.

<u>Step 1: Define</u>

<em>Identify.</em>

f(-2.4) = -1

f(-1.9) = -8

<u>Step 2: Differentiate</u>

Simply plug in the 2 coordinates into the slope formula to find slope <em>m</em>.

  1. [Derivative] Set up [Slope Formula]:                                                           \displaystyle f'(-2.4) \approx \frac{f(x_2) - f(x_1)}{x_2 - x_1}
  2. Substitute in coordinates:                                                                           \displaystyle f'(-2.4) \approx \frac{-8 - -1}{-1.9 - -2.4}
  3. Evaluate:                                                                                                       \displaystyle f'(-2.4) \approx -14

---

Learn more about derivatives: brainly.com/question/17830594

---

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

Unit: Differentiation

6 0
2 years ago
Which expression is a cube root of -1+ i sqrt 3? (multiple choice)
lara31 [8.8K]

Answer:

see below

Step-by-step explanation:

Find the magnitude   (-1)^2  + (sqrt3)^2 = m^2   m = 2

find the angle    arctan (sqrt3/-1) = 120 degrees

the cube root would then be

\sqrt[3]{2}(cos 40 + isin 40)

6 0
2 years ago
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