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Darya [45]
3 years ago
11

Multiply and simplify: y(2x+3y)(2x+3y)

Mathematics
2 answers:
vekshin13 years ago
6 0
<span>Simplifying y(2x + 3y)(2x + 3y)

Multiply (2x + 3y) * (2x + 3y) y(2x * (2x + 3y) + 3y * (2x + 3y)) y((2x * 2x + 3y * 2x) + 3y * (2x + 3y))
 
Reorder the terms: y((6xy + 4x2) + 3y * (2x + 3y)) y((6xy + 4x2) + 3y * (2x + 3y)) y(6xy + 4x2 + (2x * 3y + 3y * 3y)) y(6xy + 4x2 + (6xy + 9y2))

Reorder the terms: y(6xy + 6xy + 4x2 + 9y2)
 
Combine like terms: 6xy + 6xy = 12xy y(12xy + 4x2 + 9y2) (12xy * y + 4x2 * y + 9y2 * y) (12xy2 + 4x2y + 9y3)</span>
sladkih [1.3K]3 years ago
3 0
2xy+3y^2*2xy+3y^2
3y^2 * 3y^2
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Please help me with these
Alex Ar [27]
When we approach limits, we are finding values that are infinitesimally approaching this x-value. Essentially, we consider the approximate location that this root or limit appears. This is essential when it comes to taking Calculus, and finding the limit or rate of change of a function.

When we are attempting limits questions, there are several tests we attempt first.

1. Evaluate the limit by substituting the value of the x-value as it approaches the value (direct evaluation of a limit)
2. Rearrangement of the function, such that we can evaluate the limit.
3. (TRIGONOMETRIC PROPERTIES)
\lim_{x \to 0} (\frac{sinx}{x}) = 1
\lim_{x \to 0} (\frac{tanx}{x}) = 1
4. Using L'Hopital's Rule for indeterminate limits, such as 0/0, -infinity/infinity, or infinity/infinity.

For example:

1) \lim_{x \to 0}\frac{\sqrt{x} - 5}{x - 25}

We can do this using the first and second method.
<em>Method 1: Direct evaluation:</em>

Substitute x = 0 to the function.
\frac{\sqrt{0} - 5}{0 - 25}
= \frac{-5}{-25}
= \frac{1}{5}

<em>Method 2: Rearranging the function
</em>

We can see that x - 25 can be rewritten as: (√x - 5)(√x + 5)
By rewriting it in this form, the top will cancel with the bottom easily, and our limit comes out the same.

\lim_{x \to 0}\frac{(\sqrt{x} - 5)}{(\sqrt{x} - 5)(\sqrt{x} + 5)}
= \lim_{x \to 0}\frac{1}{(\sqrt{x} + 5)}}
= \frac{1}{5}

Every example works exactly the same way, and by remembering these criteria, every limit question should come out pretty naturally.
8 0
3 years ago
BRAINLIEST FOR RIGHT ANSWER!!
julsineya [31]

Answer: {\boxed{h(x+4)=\frac{9+3x}{8+x}}}

Concept:

In a function f(x), it represents a function in terms of x or for x. Therefore, to find the values in the function, substitute values within the parenthesis to solve.

Solve:

<u>Given information</u>

h(x)=\frac{-3+3x}{4+x}

<u>Need to find</u>

h(x+4)

<u>Substitute (x + 4) to the position of x</u>

h(x+4)=\frac{-3+3(x+4)}{4+(x+4)}

<u>Distributive property on the numerator</u>

h(x+4)=\frac{-3+3x+12}{4+(x+4)}

<u>Combine like terms</u>

h(x+4)=\frac{-3+12+3x}{4+4+x}

\boxed{h(x+4)=\frac{9+3x}{8+x}}

Hope this helps!! :)

Please let me know if you have any questions

5 0
2 years ago
Which point is a solution to the linear inequality below:
gayaneshka [121]

Answer:

A (0,0)

Step-by-step explanation:

y - 3x < 1

0 - 3(0) < 1

0 < 1

7 0
2 years ago
Find the value of x in this figure
Serjik [45]

Answer:

here question was not clear

4 0
3 years ago
Read 2 more answers
Please help fast....
Doss [256]

Answer:

Step-by-step explanation:

4 0
2 years ago
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