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3 years ago

8

You added 3 3/4 cups of flour, 1 1/4 cups of oil, and 2 2/4 cups of water to a bowl for a recipe. How many cups did you combine

in total?

1 answer:

Archy [21]3 years ago

4 0

3 3/4 + 1 1/4 + 2 2/4

3/4 + 1/4 + 2/4 = 1 1/2

3 + 1 + 2 + 1 1/2 = 7 1/2 cups total

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To calculate the area of a rectangle we multiply length by width the length of this rectangle below is 9 inches and the area is

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547

Step-by-step explanation:

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3 years ago

What is the circumference of a circle if the radius is 50 ?

Zina [86]

The appropriate formula is C = 2*pi*r.

Here, C = 2*pi*(50 units) = 100*pi units (answer)

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Triangle A″B″C″ is formed by a reflection over x = −3 and dilation by a scale factor of 3 from the origin. Which equation shows

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Answer:

see below

Step-by-step explanation:

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Then you have ...

$\overline{A''B''}=3\overline{AB}\quad\text{or}\\\\\dfrac{\overline{AB}}{\overline{A''B''}}=\dfrac{1}{3}$

The latter relation matches the second choice.

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3 years ago

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What is the multiplicative rate of change of the function shown on the graph? Express your answer in decimal form. Round to the

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In linear models there is a constant additve rate of change. For example, in the equation y = mx + b, m is the constanta additivie rate of change.

In exponential models there is a constant multiplicative rate of change.

The function of the graph seems of the exponential type, so we can expect a constant multiplicative exponential rate.

We can test that using several pair of points.

The multiplicative rate of change is calcualted in this way:

[f(a) / f(b) ] / (a - b)

Use the points given in the graph: (2, 12.5) , (1, 5) , (0, 2) , (-1, 0.8)

[12.5 / 5] / (2 - 1) = 2.5

[5 / 2] / (1 - 0) = 2.5

[2 / 0.8] / (0 - (-1) ) = 2.5

Then, do doubt, the answer is 2.5

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3 years ago

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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.

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

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

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

Unit: Limits

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1 year ago