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3 years ago

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The simplest form of the expression -2x2(x – 5) + x(2x2 – 10x) + x is

2 answers:

GalinKa [24]3 years ago

5 0

Answer:

Sounds crazy, but x

Step-by-step explanation:

Bad White [126]3 years ago

4 0

For this case you must simplify the expression respecting the rules of multiplication of mathematics.
The steps to simplify the expression are the following:
First multiply what is in the parentheses
-2x ^ 2 (x - 5) + x (2x ^ 2 - 10x) + x
-2x ^ 3 + 10x ^ 2 + 2x ^ 3 - 10x ^ 2 + x
Then add the terms that have the same exponent
(-2x ^ 3 + 2x ^ 3) + (10x ^ 2 - 10x ^ 2) + x
x
The final simplification is
x
answer
x

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3 years ago

Simplify the algebraic expression -2(x - 3) + 4(-2x + 8)

dangina [55]

The answer to this equation is :-10x+38 because
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3 years ago

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Marat540 [252]

Answer:

lol brainliest

Step-by-step explanation:

8 0

3 years ago

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Furkat [3]

Answer:

y = 8 or y = 0

Step-by-step explanation:

Solve for y over the real numbers:

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Factor y from the left hand side:

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4 0

3 years ago

For what value of constant c is the function k(x) continuous at x = 0 if k =

nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x) is continuous, we take the limit of the parameter as follows:

$\mathbf{ \lim_{x \to 0^-} k(x) = \lim_{x \to 0^+} k(x) = 0 }$

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= c }$

Provided that:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }$

Using l'Hospital's rule:

$\mathbf{\implies \lim_{x \to 0} \ \ \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}= \lim_{x \to 0} sec \ x \ tan \ x = 0}$

Therefore:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}=0 }$

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

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2 years ago