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artcher [175]
3 years ago
10

Liam is making chocolate chip cookies the recipe. calls for 1 cup of sugar for every 3 cups of flour liam has only 2 cups of flo

ur
how much sugar should liam use
Mathematics
2 answers:
LenaWriter [7]3 years ago
7 0
2/3 cup of sugar. you keep the 3 cups of flour the same, and put the 1 cup of sugar as a fraction like 3/3=1 cup. then just minus the 1 cup of flour from the top. 2/3 cups of sugar
Andrei [34K]3 years ago
4 0
Use proportion 
3 / 1 = 2/ x     where x is the number of cups of sugar

3x = 2
x = 2/3 

2/3  cups of sugar is the answer
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The average age of 6 men is 35 years and
PSYCHO15rus [73]

Answer:

39.50

42.50

Step-by-step explanation:

Age of the 2 men = sum of the ages of the 6 men - sum of the ages of the 4 men

Average = sum of ages / total number of men

35 = sum of ages / 6

sum of ages = 35 x6 = 210

32 = sum of ages / 4

32 x 4 = sum of ages

128

210 - 128 = 82

age of the younger man = x

age of the older man = 3 + x

x + 3 + x = 82

2x + 3 = 82

2x = 82 - 3

2x = 79

x = 79/2 = 39.5

older man = 39.5 + 3 = 42.50

7 0
3 years ago
Please answer! don’t forget to specify what to put for select
masha68 [24]
Angle 11 is 119
and select alternate interior angles theorem
5 0
3 years ago
Use the substitution method to solve the system of equations. Choose the
BigorU [14]

Answer:

C. (4,28)

Step-by-step explanation:

When y = 8x - 4 ; x = 4

when u substitute x in the equation

the value of y = 28

y = (8 x 4) - 4

= 32 - 4

= <u>2</u><u>8</u><u>.</u>

As coordinates it would be (x,y) = <u>( 4,28 )</u>

3 0
2 years ago
Read 2 more answers
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]:

  1. f(x) = cxⁿ
  2. 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:

  1. [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}
  2. 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:

  1. [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}
  2. [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}
  3. [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}
  4. 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
Good morning can y’all help me
Ulleksa [173]

Answer:

rate of change

Step-by-step explanation:

8 0
3 years ago
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