#1**+1 **

a) In order to factor using this method, let's try and identify the GCF first. 9 is the greatest common factor between 36 and 81. a^4 is the greatest common factor between the a's, and b^10 is the factor for the b's. Let's factor it out!

\(36a^4b^{10}-81a^{16}b^{20}\) | Factor out the GCF, \(9a^4b^{10}\), like I described earlier. |

\(9a^4b^{10}\left(4-9a^{12}b^{10}\right)\) | Don't stop here, though! Notice that the resulting binomial is a difference of squares. |

\(9a^4b^{10}\left(2+3a^6b^5\right)\left(2-3a^6b^5\right)\) | |

b) The beginning binomial is a difference of squares to begin with, so it is possible to start with this first!

\(36a^4b^{10}-81a^{16}b^{20}\) | Let's do this approach this time! |

\(\left(6a^2b^5+9a^8b^{10}\right)\left(6a^2b^5-9a^8b^{10}\right)\) | Don't stop yet! Both binomials have their own GCF's! |

\(3a^2b^5\left(2+3a^6b^5\right)*3a^2b^5\left(2-3a^6b^5\right)\) | Combine the multiplication. |

\(9a^4b^{10}\left(2+3a^6b^5\right)\left(2-3a^6b^5\right)\) | |

Well, these are the two techniques.

TheXSquaredFactor
Feb 24, 2018

#1**+1 **

8) This one will require the formula that yields the volume of a cylinder. \(V_{\text{cylinder}}=\pi r^2h\). We can manipulate this formula so that we can find any missing information such as the height, in this case.

\(V_{\text{cylinder}}=\pi r^2h\) | We know what the volume is, and we know the height, so finding the radius is simply a matter of isolating the variable. | ||

\(27143=15\pi r^2\) | Divide by 15 pi first. | ||

\(\frac{27143}{15\pi}=r^2\) | Take the square root of both sides. | ||

\(|r|=\sqrt{\frac{27143}{15\pi}}\) | The absolute value splits the answer into two possibilities. | ||

| In the context of geometry, negative side lengths are nonsensical, so let's just reject the answer now. | ||

\(r=\sqrt{\frac{27143}{15\pi}}\approx 24\text{m}\) | The radius is a one-dimensional part of a cylinder, so the units should be in one dimension, too. | ||

9) If the height of the un-consumed soup was 8 centimeters tall and 3-centimeters-worth of soup is consumed, then 5 centimeters of soup remains. We already know the radius of this soup can (that I assume is cylinder-shaped despite not being explicitly stated), so we can determine the volume.

\(V_{\text{cylinder}}=\pi r^2h\) | Plug in the known values. |

\(V_{\text{cylinder}}=\pi*12^2*5\) | Now, combine like terms. |

\(V_{\text{cylinder}}=720\pi\approx 2262\text{cm}^3\) | Volume is always expressed as a cubic unit. |

10) If the town park enlarges its area by a factor of 5, then both dimensions of the park are affected by this scale factor. For example, if we assume, for the sake of understanding, that the park is perfectly rectangular with dimensions 3yd by 100yd, then both dimensions (the length and the width) would be affected by this scale factor. This means that, on area, the scale factor actually affects the area by its square, or 25 in this case.

\(300\text{yd}^2*25=7500\text{yd}^2\)

TheXSquaredFactor
Feb 24, 2018

#1**+1 **

The problem is easier than I first thought.

By the given information, we know that there are three right-angled triangles in the diagram. We know that \(m\angle AEB=m\angle BEC=m\angle CED= 60^{\circ}\). We can use this information to determine that every right triangle is also a 30-60-90 triangle. We also know that \(AE=24\).

A 30-60-90 triangle is a special kind of right triangle where the ratio of the side lengths are \(1:\sqrt{3}:2\). \(\overline{AE}\) is the longest side length because it is the hypotenuse of the largest right triangle in the diagram. \(\overline{BE}\) is the shortest side length of \(\triangle ABE\) because this side is opposite the smallest angle. We mentioned earlier what the ratio of the side lengths are, so we can determine the length of \(\overline{BE}\) without doing anything too computationally demanding.

\(\frac{BE}{AE}=\frac{1}{2}\) | We already know the ratio of the side lengths of a 30-60-90 triangle, so we can apply this relationship and create a proportion. We already know what AE equals, so let's fill that in. |

\(\frac{BE}{24}=\frac{1}{2}\) | In order to solve a proportion, simply cross multiply. |

\(2BE=24\) | Divide by 2 on both sides to determine the unknown length of the side. |

\(BE=12\) | |

Of course, the ultimate goal is to figure out the length of \(\overline{CE}\). If you look at \(\triangle BCE\), carefully, you will notice that we are in an identical situation to when we solved for \(BE\). Notice that \(\overline{BE}\) is the hypotenuse of this triangle, and \(\overline{CE}\) is the shortest side length since it is opposite the 30º angle. We can use the same \(1:\sqrt{3}:2\) relationship of the side lengths to find the missing length.

\(\frac{CE}{BE}=\frac{1}{2}\) | Just like before, we know what the value of BE is, so let's plug it in! |

\(\frac{CE}{12}=\frac{1}{2}\) | Just like before, cross multiplying is the way to go! |

\(2CE=12\) | Divide by 2 on both sides to solve this problem. |

\(CE=6\) | |

TheXSquaredFactor
Feb 24, 2018

#1**0 **

This question requires one to understand certain features of a rational function in order to determine the horizontal asymptote.

#1) The degree of the numerator is 6, and the degree of the denominator is 5. Since the degree of the numerator exceeds the degree of the denominator, no horizontal asymptote exists for the first function.

#2) There is a general process to graphing rational functions.

**1**) **Factor the numerator and denominator completely, if possible**

In this case, no factoring can be done to either the numerator or the denominator. If it were possible, the process would expose any hidden common factors.

**2) Identify any Holes**

We can essentially skip this step; holes are generated when a common factor between the numerator and denominator exists. We would have identified the common factor in the previous step.

**3) ****Identify any Zeros**

Setting the numerator equal to zero allows one to identify the zeros. The numerator of this rational function is not complex by any means, so it is relatively easy to find the zero.

\(-3x+5=0\) | |

\(-3x=-5\) | |

\(x=\frac{-5}{-3}=\frac{5}{3}\) | |

Since we are solving for a zero, the y-coordinate equals zero; thus, there is a zero at \(\left(\frac{5}{3},0\right)\).

**4) ****Identify any Asymptotes**

Of course, there are three types of asymptotes (vertical, horizontal, and oblique), so we need to be sure to take all of them into account, if they exist.

Setting the denominator equal to zero reveals the vertical asymptote.

\(-5x+2=0\) | |

\(-5x=-2\) | |

\(x=\frac{-2}{-5}=\frac{2}{5}\) | |

There is a vertical asymptote at x=2/5 |

The horizontal asymptotes can be determined by the degree of both the numerator and denominator. In this case, the degree of the numerator and denominator are equal, so you would divide the leading coefficients of the numerator and denominator.

The horizontal asymptote exists at \(y=\frac{-3}{-5}=\frac{3}{5}\)

For rational functions, it is impossible that an oblique asymptote exists if a horizontal asymptote does, so there is no oblique asymptote.

**5) Plot**** any Information Determined Previously**

We know where a zero exists already (at \(\left(\frac{3}{5},0\right)\)), so we might as well plot it.

Plotting asymptotes are also important; they tell where functions approach, so the function does not cross asymptotes. Be careful, though! For rational functions, a function will never pass a vertical asymptote, but it can pass a horizontal or oblique asymptote. In this case, though, the function will not pass through any asymptotes.

**6) Create a Table of Values**

If, after this process, you are still unsure about how a graph behaves, creating a table of values might be your best solution. Be strategic about it, though! Plot points on all sides of vertical asymptotes to better understand the behavior.

TheXSquaredFactor
Feb 22, 2018

#3**+1 **

sii1lver, you have already asked this question. Two more explanations for this question are located at this link: https://web2.0calc.com/questions/i-also-need-help-on-questions-3-7-and-8-too-these#r5

Yes, the individual answers are not identical, but it is possible to attribute the small inaccuracy to rounding.

TheXSquaredFactor
Feb 18, 2018

#6**0 **

I am willing to share another method for the eighth question

#8)

The method I default to uses the following conversion. I will create another table:

\(46g\text{Cl}_2\) | \(1\text{molCl}_2\) | \(22.4L\text{Cl}_2\text{@STP}\) |

\(?g\text{Cl}_2\) | \(1\text{molCl}_2\) |

There is only one unknown here! We only need to find the number of grams that equals the number of moles of Cl_{2}. There are 35.45 grams of Cl per mole, according to the trusty periodic table. Cl_{2} has double the number of molecules as Cl, so \(1\text{molCl}_2=35.45g*2=70.9g\). In case you are unaware, "STP" is shorthand for standard temperature and pressure.

\(46g\text{Cl}_2*\frac{1\text{molCl}_2}{70.9g\text{Cl}_2}*\frac{22.4L\text{Cl}_2\text{@STP}}{1\text{molCl}_2}\) | Cancel out all the units. |

\(46*\frac{1}{70.9}*\frac{22.4L\text{Cl}_2\text{@STP}}{1}\) | The rest is a calculator's job. Yet again, the guest answer and my answer are close. |

\(14.53L\text{CL}_2\) |

TheXSquaredFactor
Feb 16, 2018

#5**+2 **

This method is somewhat different than the guest's because it does not need to consider the amount of Cl_{2} reacted. The general conversion strategy for this particular procedure goes as follows. The table displays the general layout, and all unknowns are marked with question marks:

\(\text{grams}\rightarrow\text{moles}\rightarrow\text{moles}\rightarrow\text{grams}\)

\(16.2g\text{H}_2\) | \(1\text{molH}_2\) | \(?\text{molHCl}\) | \(?g\text{HCl}\) |

\(?g\text{H}_2\) | \(?\text{molH}_2\) | \(1\text{molHCl}\) |

Let's travel through this table column by column. We start with the given information, 16.2gH_{2}, and the eventual goal is to perform a series of conversions. The first column of the table is already finished.

The second column asks the following question: How many grams of H_{2 }are in one mole of H_{2}? In order to answer this question, we have to reference the indispensable periodic table of the elements. I generally use https://www.ptable.com/

as an electronic version of the table.

The atomic mass of an element is also the molar mass represented in grams, so H has a molar mass of 1.008g. However, realize that we are finding the molar mass of H_{2}. The subscript indicates that there are two hydrogen molecules, so double the original molar mass, 1.008g, to obtain the molar mass of H_{2}. \(1\text{molH}_2=1.008g*2=2.016g\).\(\)

The third column is quite a simple step, actually. It compares the molar ratio of the two molecules in question, HCl and H_{2}, in this case. Determining this information requires some basic knowledge of a balanced equation. In the given chemical reaction, it is possible to perceive it in the following sense: One molecule of H_{2} reacts and yields two molecules of HCl. The number of molecules contained in a mole equals Avagadro's constant, or \(1\text{mol}=6.02*10^{23}\text{ molecules}\). If you continue this logic, the original balanced equation indicates \(1\text{molH}_2=2\text{molHCl}\).

The procedure for the fourth column is identical to the procedure for the second column. How many grams of HCl equals one mole of HCl? Because HCl is a compound, the combined mass of the elements equals its molar mass. As aforementioned, H has a mass of 1.008g per mole. Cl, according to the trusty periodic table, has a mass of 35.45g per mole. Therefore, \(1\text{molHCl}=1.008g+35.45g=36.458g\).

After all this work, we have finally determined all the missing values in the original conversion that I suggested earlier. The table now looks complete.

\(16.2g\text{H}_2\) | \(1\text{molH}_2\) | \(2\text{molHCl}\) | \(36.458g\text{HCl}\) |

\(2.016g\text{H}_2\) | \(1\text{molH}_2\) | \(1\text{molHCl}\) |

This table is really a fancy representation of three ratios.

\(16.2g\text{H}_2* \frac{1\text{molH}_2}{2.016g\text{H}_2}* \frac{2\text{molHCl}}{1\text{molH}_2}* \frac{36.458g\text{HCl}}{1\text{molHCl}}\) | First and foremost, let's cancel out all the common units. Doing this shows that the only unit remaining is the desired unit. |

\(16.2* \frac{1}{2.016}* \frac{2}{1}* \frac{36.458g\text{HCl}}{1}\) | Now it is a matter of simplifying. When I input this entire expression into the calculator, I get an answer close to what the guest got. |

\(586.0g\text{HCl}\) | |

I default to this method because it removes the need to approximate halfway through the calculation.

TheXSquaredFactor
Feb 16, 2018

#1**+1 **

I think I can answer #1

The diagram below meets all the given criteria. It is a quadrilateral with the given side lengths. Since \(\overline{AC}\) is the hypotenuse of \(\triangle ABC\), we can use the Pythagorean Theorem to find the length of the missing side length. Of course, this triangle satisfies the most famous pythagorean triple, a 3-4-5 right triangle, so \(AC=5\).

Yet again, \(\triangle DAC\) is in a similar situation. It is a 5-12-13 triangle, which is another Pythagorean triple. Because of this, \(m\angle DAC=90^\circ\).

I can break up the area of the original quadrilateral into two smaller parts: the 3-4-5 right triangle and the 5-12-13 right triangle.

Using the formula \(A_{\triangle}=\frac{1}{2}bh\), one can identify the areas of both triangles. This process is even simpler since the height of both triangles is also the perpendicular height.

\(A_1=\frac{1}{2}*3*4\\ A_1=2*3\\ A_1=6\) | \(A_2=\frac{1}{2}*5*12\\ A_2=6*5\\ A_2=30\) |

Now, just add the areas together.

\(A_1+A_2=A_{\text{total}}\\ 6+30=36\text{units}^2\)

Many of the questions seem to suggest that there is a complementary diagram or some missing information (2, 3, 5 and 6)

TheXSquaredFactor
Feb 15, 2018

#2**+1 **

#10, why not?

According to the original relation, \(f(x)=\sqrt[3]{x-4}\). There are a few steps that need to be done in order to find the inverse.

#1) Convert to y=-notation. I find that it is easier when working with this notation. This is quite a simple step, wouldn't you agree?

\(f(x)=\sqrt[3]{x-4}\Rightarrow y=\sqrt[3]{x-4}\)

#2) Now, replace all instances of a "y" with an "x," and replace all instances of an "x" with a "y." This is not a difficult step, either, as you might imagine.

\(y=\sqrt[3]{x-4}\Rightarrow x=\sqrt[3]{y-4}\)

#3) Solve for y. This step can range in difficulty. In this case, it is quite a simple step.

\(x=\sqrt[3]{y-4}\) | Cube both sides. |

\(x^3=y-4\) | Add 4 to both sides to isolate y. |

\(y=x^3+4\) | |

#4) Convert back to function notation since the original problem was given in function notation. This is quite simple, too.

\(y=x^3+4\Rightarrow f^{-1}(x)=x^3+4\)

#5) Correspond your answer with the answer choices given!

TheXSquaredFactor
Feb 13, 2018

#1**+1 **

To solve for a variable, in this case, simpl means to isolate it. That is the goal of solving for a variable.

1)

\(C=K\left(\frac{Rr}{R-r}\right)\) | Let's first multiply both sides by (R-r). This will eliminate the denominator, which will ease the process of solving for "K." |

\(C(R-r)=KRr\) | Divide by Rr to isolate K. |

\(K=\frac{C(R-r)}{Rr}\\\) | You could distribute, if you so desired, but it is not necessary. This just creates more work for yourself, anyway. This is the final answer. |

2) This one will be harder since two "R's" exist in the original equation. This generally signals for the use of grouping.

\(C=K\left(\frac{Rr}{R-r}\right)\) | Let's do the exact same thing as before; multiply by R-r. |

\(C(R-r)=KRr\) | Let's distribute the C into the binomial. This will allow us to have two terms with "R's" |

\(CR-Cr=KRr\) | Let's subtract KRr from both sides. Let's add Cr to both means meanwhile. |

\(CR-KRr=Cr\) | Using grouping, we can change this into one "R." |

\(R(C-Kr)=Cr\) | Now, divide by C-Kr to isolate "R." |

\(R=\frac{Cr}{C-Kr}\) | This is finished because we have isolated the R. |

3) We will have to utilize the same technique in number 2 in order to isolate "f."

\(F=\frac{fg}{f+g-d}\) | Let's multiply the denominator on both sides. This will get rid of any and all pesky denominators in this equation, |

\(F(f+g-d)=fg\) | There is a "g" in the trinomial, so we will have to expand that part. |

\(Ff+Fg-Fd=fg\) | Move all terms with a factor of f on the left hand side. Any other terms are moved to the right hand side. |

\(Ff-fg=-Fg+Fd\) | Factor out an "f" from both terms from the left hand side. |

\(f(F-g)=-Fg+Fd\) | Finally, divide by F-g to isolate the wanted variable completey. |

\(f=\frac{-Fg+Fd}{F-g}\) | No more simplification is possible here. Leave the fraction as is. |

TheXSquaredFactor
Feb 13, 2018

#5**+1 **

For the second question, it is possible to utilize something known as the Pythagorean Inequality Theorem. The theorem actually has two parts, but the relevant portion for this particular problem is that if the square of a triangle's longest side length is greater than the the sum of the squares of the shorter side lengths, then the triangle is obtuse. This wording is quite verbose, I know, so I decided to create a visual aide as a reference for you!

Let's first try and identify what the longest side length is of this particular triangle is. We know that 16 and 21 are already defined lengths, but we also have the unknown side length, which is labeled as *s*. I will follow the visual aide above, and I will consider the longest side length to be of length "c." In the visual aide above, the shorter side lengths are labeled "a" and "b."

It is easy to exclude the side with length 16 as the longest side length since we know that there exists a side with length 21, which is longer. However, beyond this, there are two possibilities. The possibility is that:

- The side with length
*s*is the longest side. - The side with length 21 is the longest side.

Of course, we do not know which one is the longest side, so we will just have to assume both cases. Let's begin with assuming that "s" is the longest side, or our "c."

\(c^2>a^2+b^2\) | "s" will be the longest side, so plug it into "c." Plug the other two side lengths in any order. | ||

\(s^2>21^2+16^2\) | Now, solve for "s" to see what the possibilities are. | ||

\(s^2>441+256\) | |||

\(s^2>697\) | Take the square root of both sides. This results in an absolute value inequality. | ||

\(|s|>\sqrt{697}\) | Of course, the absolute value splits the equation into two parts. | ||

| Of course, let's remember that "s" is an integer side length of a triangle, so "s" must be positive. We can reject the second inequality because those values include ones less tha zero. | ||

\(s>\sqrt{697}\) |

The above algebra only solved the first case where we assumed that "s" is the longest side. Now, let's assume that the side with length 21 is the indeed the longest.

\(21^2>16^2+s^2\) | Let's solve for "s" here by simplifying first and foremost. | ||

\(441>256+s^2\) | Subtract 256 from both sides. | ||

\(185>s^2\\ s^2<185\) | For me, I can interpret the inequality better when the variable is placed on the left hand side of the equation. Take the square root of both sides. | ||

\(|s|<\sqrt{185}\) | Now, solve the absolute value inequality. | ||

| Unlike the previous inequality, we can combine this into one compound inequality. | ||

-√185 < s < √185 | Of course, yet again, "s" is a side length, so it should be greater than zero. | ||

0 < s < √185 |

Of course, let's take clean this up. We know, by the given information, that we only care about integer solutions, so let's calculate the radicals.We now have set restrictions for "s." \(\)

√697 ≈ 26.40

√185 ≈ 13.60

We can then make the restrictions the following

We are not done yet, though! We have to take into account the side length restriction that Cpill figured out in part a! He found that 5 < s < 37

Therefore, integer sides with lengths 6 to 13 or sides with lengths 27 to 36 are possible. We can find the total number of integer solutions by figuring out how many integers are in this range.

From [6,13], there are 8 integers

From [27,36], there are 10 integers

Altogether, there are 18 integer solutions.

*Unfortunately, LaTeX appears to be funky when compound inequalities are introduced, so I have resorted to a different method to portray them.

TheXSquaredFactor
Feb 13, 2018