Ask question

Ask question

All categories

2 years ago

8

A recipe calls for 6 ounces of water. If there are 16 tablespoons in one cup. and two tablespoons in once ounce. how many cups o

f water do you need

1 answer:

Gnom [1K]2 years ago

8 0

Answer:

3/4 cup

Step-by-step explanation:

You might be interested in

A company has recently been hiring new employees. Today the company has 34% more employees than it did a year ago. If there are

Anton [14]

Answer:

22110 employees.

Step-by-step explanation:

33500 decrease 34% =

33500 × (1 - 34%) = 33500 × (1 - 0.34) = 22110

5 0

3 years ago

Read 2 more answers

PLSSSSSSS HELPPPPPPP I WILL GIVE BRAINLIESTTTTTTTTTT!!!!!!!!!!!!!!!!!!!!!PLSSSSSSS HELPPPPPPP I WILL GIVE BRAINLIESTTTTTTTTTT!!!

Arisa [49]

Answer: y^2/x

Step-by-step explanation:

8 0

2 years ago

What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?

scoray [572]

Answer:

<u /> $\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

<u>Calculus</u>

Limits

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

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$
Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify given limit</em>.

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

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

Let's start out by <em>directly</em> evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}$

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

[Limit] Apply Limit Rule [L' Hopital's Rule]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}$
[Limit] Differentiate [Derivative Rules and Properties]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$

∴ we have <em>evaluated</em> the given limit.

___

Learn more about limits: brainly.com/question/27807253

Learn more about Calculus: brainly.com/question/27805589

___

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

Unit: Limits

3 0

1 year ago

One of the events in the Winter Olympics is the Men's 500-meter Speed Skating.

poizon [28]

Answer:

Mean: $\bar x = 40.61$

$Median =40.2$

$Mode = 43.4$

Step-by-step explanation:

Given

See attachment for data

Solving (a): The mean

Mean is calculated as:

$\bar x = \frac{\sum x}{n}$

From the attached:

$n = 14$

So, the mean is:

$\bar x = \frac{43.4+43.4+43.4+43.1+43.2+40.2+40.2+40.1+40.3+39.44+39.17+38.03+38.19+36.45}{14}$

$\bar x = \frac{568.58}{14}$

$\bar x = 40.61$

Solving (b): The median

$n = 14$ ; this is an even number. So, the median is:

$Median = \frac{1}{2}(n+1)$

$Median = \frac{1}{2}(14+1)$

$Median = \frac{1}{2}(15)$

$Median = 7.5th$

This implies that the median is the average of the 7th and 8th item.

Next, is to order the data (in ascending order): <em>36.45, 38.03, 38.19, 39.17, 39.44, 40.1, 40.2, 40.2, 40.3, 43.1, 43.2, 43.4, 43.4, 43.4.</em>

The 7th and 8th items are: 40.2 and 40.2

The median is:

$Median = \frac{1}{2}(40.2 + 40.2)$

$Median = \frac{1}{2}*80.4$

$Median =40.2$

Solving (c): The mode

43.4 has the highest number of occurrence.

So:

$Mode = 43.4$

8 0

3 years ago

Match the rate - - .75 -- with corresponding time of year

vovikov84 [41]

Answer: C. Nine Months, 9/12 = 9÷12 = 0.75

Step-by-step explanation:

Hi, to answer this question we have to convert the fraction into decimals and compare:

A. Three Months, 1/4 = 1÷4 = 0.25

B. Six Months, 6/12 = 6÷12 =0.5

C. Nine Months, 9/12 = 9÷12 = 0.75

D. 1 Month, 1/12 =1÷12 = 0.083333

C. Nine Months, 9/12 = 9÷12 = 0.75

Feel free to ask for more if needed or if you did not understand something.

6 0

3 years ago