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3 years ago

Is the expression completely factored: (5 - x)(6 - 5x)

Mathematics

1 answer:

Naya [18.7K]3 years ago

7 0

Answer:

5x² - 31x + 30

Step-by-step explanation:

(5 - x)(6 - 5x)

= 30 - 25x - 6x + 5x²

= 30 - 31x + 5x²

= 5x² - 31x + 30

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Multiple choice please please answer

lakkis [162]

Answer:

$\purple { \boxed{x = 25 \times \sin \: 39 \degree}}$

Step-by-step explanation:

$\sin \: 39 \degree = \frac{x}{25} \\ 25 \times \sin \: 39 \degree = x \\ \huge \red{ \boxed{x = 25 \times \sin \: 39 \degree}}$

4 0

3 years ago

Triangle JKL is dilated from the origin at a scale factor of 10 to create triangle J′K′L′.Select the option that completes the s

Aneli [31]

Given: Triangle JKL is dilated from the origin at a scale factor of 10 to create triangle J′K′L′.

Required: Correct option that complete the statement.

Explanation:

Now, triangle is dilated from the origin at a scale factor of 10.

So, the sides will become larger, but angles will remain the same.

Hence, the triangles will be similar as sides will be in proption (SSS rule).

So, The triangles are similar because their corresponding sides are proptional and their corresponding angles are congurent.

Final Answer: Second group is correct answer.

5 0

1 year ago

If the area of a triangle is 48 cm2 and the base of the

makvit [3.9K]

Answer:

Step-by-step explanation:

Formula for height of triangle

A÷BX2

In this case, 48÷12=4×2=8

Hope you understand ☺️

4 0

3 years ago

The function C = x - 2y is maximized at the vertex point of the feasible region at (8, 2). What is the maximum value?

lubasha [3.4K]

The required maximum value of the function C = x - 2y is 4.

Given that,
The function C = x - 2y is maximized at the vertex point of the feasible region at (8, 2). What is the maximum value is to be determined.

<h3>What is the equation?</h3>

The equation is the relationship between variables and represented as y =ax +m is an example of a polynomial equation.

Here,
Function C = x - 2y
At the vertex point of the feasible region at (8, 2)
C = 8 - 2 *2
C= 4

Thus, the required maximum value of the function C = x - 2y is 4.

Learn more about equation here:

brainly.com/question/10413253

#SPJ1

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1 year ago

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

scoray [572]

Answer:

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

General Formulas and Concepts:

Calculus

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:

Step 1: Define

Identify given limit.

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

Step 2: Find Limit

Let's start out by directly 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 evaluated 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

3 0

2 years ago