Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

Solve for [tex]\( x \)[/tex].

[tex]\[ 3x = 6x - 2 \][/tex]


Format the following question or task so that it is easier to read.
Fix any grammar or spelling errors.
Remove phrases that are not part of the question.
Do not remove or change LaTeX formatting.
Do not change or remove [tex] [/tex] tags.
If the question is nonsense, rewrite it so that it makes sense.
-----
2
TIME REMAINII
The volume of a rectangular prism is [tex]$\left(x^4+4 x^3+3 x^2+8 x+4\right)$[/tex], and the area of its base is [tex]$\left(x^3+3 x^2+8\right)$[/tex]. If the volum of a rectangular prism is the product of its base area and height, what is the height of the prism?
[tex]$x+1-\frac{4}{x^4+4 x^3+3 x^2+8 x+4}$[/tex]
[tex]$x+1+\frac{4}{x^4+4 x^3+3 x^2+8 x+4}$[/tex]
[tex]$x+1-\frac{4}{x^3+3 x^2+8}$[/tex]
[tex]$x+1+\frac{4}{x^3+3 x^2+8}$[/tex]
-----

Response:

The volume of a rectangular prism is [tex]\( x^4+4x^3+3x^2+8x+4 \)[/tex], and the area of its base is [tex]\( x^3+3x^2+8 \)[/tex].

If the volume of a rectangular prism is the product of its base area and height, what is the height of the prism?

A. [tex]\( x+1-\frac{4}{x^4+4x^3+3x^2+8x+4} \)[/tex]

B. [tex]\( x+1+\frac{4}{x^4+4x^3+3x^2+8x+4} \)[/tex]

C. [tex]\( x+1-\frac{4}{x^3+3x^2+8} \)[/tex]

D. [tex]\( x+1+\frac{4}{x^3+3x^2+8} \)[/tex]


Sagot :

To solve the problem of finding the height of the rectangular prism, we start by applying the formula for the volume of a rectangular prism, which is given by the product of its base area and height:

[tex]\[ \text{Volume} = \text{Base Area} \times \text{Height} \][/tex]

Given:
- The volume of the rectangular prism is [tex]\( V = x^4 + 4x^3 + 3x^2 + 8x + 4 \)[/tex]
- The area of the base of the rectangular prism is [tex]\( A = x^3 + 3x^2 + 8 \)[/tex]

We need to determine the height [tex]\( h \)[/tex] of the prism. We can derive the height by rearranging the volume formula to solve for [tex]\( h \)[/tex]:

[tex]\[ h = \frac{V}{A} \][/tex]

Substitute the expressions for the volume and base area:

[tex]\[ h = \frac{x^4 + 4x^3 + 3x^2 + 8x + 4}{x^3 + 3x^2 + 8} \][/tex]

Next, we simplify the expression [tex]\( \frac{x^4 + 4x^3 + 3x^2 + 8x + 4}{x^3 + 3x^2 + 8} \)[/tex]:

To simplify this rational function, consider polynomial long division:

1. The leading term of the numerator [tex]\( x^4 \)[/tex] divided by the leading term of the denominator [tex]\( x^3 \)[/tex]:
[tex]\[ x \][/tex]

2. Multiply the entire divisor [tex]\( x^3 + 3x^2 + 8 \)[/tex] by this quotient:
[tex]\[ x(x^3 + 3x^2 + 8) = x^4 + 3x^3 + 8x \][/tex]

3. Subtract this product from the numerator:
[tex]\[ (x^4 + 4x^3 + 3x^2 + 8x + 4) - (x^4 + 3x^3 + 8x) = x^4 + 4x^3 + 3x^2 + 8x + 4 - x^4 - 3x^3 - 8x = x^3 + 3x^2 + 4 \][/tex]

4. The new expression becomes:
[tex]\[ \frac{x^3 + 3x^2 + 4}{x^3 + 3x^2 + 8} \][/tex]

We notice that by performing polynomial division, we have:
[tex]\[ x^3 + 3x^2 + 4 \][/tex]

This shows a quotient simplification leading to the final result, combined with the original leading term quotient [tex]\( x \)[/tex]:

Now we test against the options to match it properly.

Upon simplifying and matching with the given options:
- [tex]\( x+1-\frac{4}{x^3+3 x^2+8} \)[/tex]
- [tex]\( x+1+\frac{4}{x^3+3 x^2+8} \)[/tex]
- [tex]\( x+1-\frac{4}{x^4+4 x^3+3 x^2+8 x+4} \)[/tex]
- [tex]\( x+1+\frac{4}{x^4+4 x^3+3 x^2+8 x+4} \)[/tex]

We find that:

[tex]\[ \frac{x^4 + 4x^3 + 3x^2 + 8x + 4}{x^3 + 3x^2 + 8} = x + 1 - \frac{4}{x^3 + 3x^2 + 8} \][/tex]

Thus, the height of the prism is:
[tex]\[ \boxed{3} \][/tex]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.