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2 years ago

15

Mrs.Brown gives two tests to her class. 30% of the class passed both tests and 45% of the class passed the first test. What perc

ent passed the second test given that they passed the first test?

1 answer:

viktelen [127]2 years ago

7 0

Answer:

25%

Step-by-step explanation:

100%-30%=70%

70%-45%=25%

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Given: m∠p =47BD∥AC <br> What is the m∠m?<br><br> Put only the number. Do not label your answer.

Anettt [7]

Answer: $43\°$

Step-by-step explanation:

Observe the given figure.

You can identify that the measure of the angle "B" is:

$m\angle B=90\°$

Knowing this, you can conclude that the angle "p" and the angle "m" are Complementary angles, which means that the sum of their measures is 90 degrees. Then, you can write the following equation:

$m\angle p+m\angle m=90\°$ [Equation 1]

You know that:

$m\angle p=47\°$

Then you can substitute it into the [Equation 1] and then solve for $m\angle m$ . Therefore, you get:

$47\°+m\angle m=90\°\\\\m\angle m=90\°-47\°\\\\m\angle m=43\°$

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3 years ago

Write a polynomial that fits the description:

8090 [49]

4x^5 + 3x + 1...the ^5 makes it a fifth degree....and the leading coefficient is 4

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3 years ago

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A function is represented by the graph.

Morgarella [4.7K]

Answer:

see the explanation

Step-by-step explanation:

we know that

The y-intercept is the value of the function y when the value of x is equal to zero

Part 1) we have

$y=5x-6$

For x=0

substitute in the linear equation and solve for y

$y=5(0)-6$

$y=-6$

therefore

The y-intercept is the point (0,-6)

Part 2) Find the y-intercept of the function represented in the graph

Looking at the graph

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The value of y is equal to y=1

therefore

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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.

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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

David considered moving even closer to his workplace. He claims that if he had done so, the percent of decrease would have been

krok68 [10]

This is completely wrong for logical and mathematical reasons, but for the sake of time I will go over the mathematical reason.

Because all percentages are out of 100, nothing can physically be over 100%, though it is used frequently in hyperbolic sentences and exaggerated statements.

Hope I helped and good luck!

3 0

3 years ago

Read 2 more answers