PLEASE HELP ASAP!!!!

Given cos A = .23 find angle A in radians. Round your answer to the nearest hundredth.

irinina [24]

Use a calculator and change DEG (degres) to RAD (radians).

8 0

3 years ago

A stack of round crackers are in the shape of an oblique cylinder. The diameter of each cracker is 6 centimeters and the height

tekilochka [14]

Answer: 294cm

Step-by-step explanation:

The volume of a oblique cylinder is given by:

Volume= πr^2h

where

π = 3.14

r = radius

h = height

Radius= diameter/2 = 6/2 = 3

We have all the values, so we put it in the formulae so we can get the answer.

Volume= πr^2h

= 3.14 × 3 × 3 × (0.4 × 26)

= 294cm

8 0

3 years ago

Can you find the limits of this

Pavel [41]

Answer:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{-3}{8}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Constant]: $\displaystyle \lim_{x \to c} b = b$

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

Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

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:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

We are given the following limit:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16}$

Let's substitute in x = -2 using the limit rule:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{(-2)^3 + 8}{(-2)^4 - 16}$

Evaluating this, we arrive at an indeterminate form:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{0}{0}$

Since we have an indeterminate form, let's use L'Hopital's Rule. Differentiate both the numerator and denominator respectively:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \lim_{x \to -2} \frac{3x^2}{4x^3}$

Substitute in x = -2 using the limit rule:

$\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{3(-2)^2}{4(-2)^3}$

Evaluating this, we get:

$\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{-3}{8}$

And we have our answer.

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

Unit: Limits

6 0

2 years ago

Which expression correctly represents “six more than the quotient of three and a number, decreased by eight”?

Neporo4naja [7]

Answer:

C

Step-by-step explanation:

Because yeahhhh

7 0

2 years ago

Read 2 more answers

Pls show all work. I will mark BRAINLIEST. Plsssss thank you and will the to give as many points. Answer has to be right. Due 11

dimulka [17.4K]

Answer: N0. There are not enough teacher signatures.

Step-by-step explanation:

Total signatures needed: $\frac{1}{8}$ x 1200 = 150

Total signatures received: 150

********************************************************************

Teacher signatures needed: $\frac{1}{25}$ x 150 = 6

Teacher signatures received: $\frac{2}{75}$ x 150 = 4

Side note: If there are $\frac{73}{75}$ student signatures then there are $\frac{2}{75}$ teacher signatures, because the student plus teacher signatures must equal 100%

8 0

3 years ago

Read 2 more answers