+2441 AngelRay, my strategy here would be to convert all inequalities to slope-intercept form (\(y=mx+b\)). Converting to slope-intercept form is useful because the form gives one information about how a graph should look. I can, then, compare this information with the four graphs given.
| \(x-y<-1\) | Add x to both sides. |
| \(-y<-x-1\) | Divide by -1 on both sides. Doing this causes an equality signage change. |
| \(y>x+1\) | |
The point-intercept form gives us some information about the corresponding graph of this equation. I will name two of these.
Every graph appears to have the correct slope, but the coordinate of the y-intercept changes from graph to graph. Only the first and last graphs have a y-intercept of \((1,0)\) . Therefore, we have already eliminated two graphs from the list. Notice that the inequality would result in a dashed line. Only the last graph accounts for this technicality, so the bottom-right one is the correct graph.
+2441 #1)
Using the similarity statement, we can find \(YR\) by generating a proportion.
| \(\frac{DZ}{LZ}=\frac{MR}{YR}\) | There are a few options here. This proportion is the one I chose to set up for this particular problem. Substitute in the known side lengths and solve for the missing one. |
| \(\frac{18}{21}=\frac{28.8}{YR}\) | It is generally wise to simplify any fractions completely before one cross multiplies. Taking this precaution beforehand can ensure that the computation does not get out of hand. |
| \(\frac{6}{7}=\frac{28.8}{YR}\) | Now we can cross multiply. |
| \(6YR=201.6\) | |
| \(YR=33.6\text{in}\) | Do not forget the units! |
| \(YR=34\text{in}\) | The question asks that the answer is rounded to the nearest whole inch, so I complied. |
#2)
This question becomes simple once you know the formula for the volume of a pyramid: \(V_{\text{pyramid}}=\frac{1}{3} lwh\)
| \(V_{\text{pyramid}}=\frac{1}{3}lw h\) | Of course, we must look at the given information; the figure is a square pyramid, so the side length of the bases is equivalent. |
| \(V_{\text{pyramid}}=\frac{1}{3}*183*183*110\) | 183 is divisible by 3, so we can reduce that portion now. |
| \(V_{\text{pyramid}}=61*183*110\) | |
| \(V_{\text{pyramid}}=1227930\text{m}^3\) | |
#3)
The fill-in-the-blank questions are really just testing one's knowledge of the individual formulas.
\(V_{\text{cylinder}}=\hspace{3mm}\pi r^2 h\\ V_{\text{cone}}\hspace{5mm}=\frac{1}{3} \pi r^2 h\)
When you place the formulas side by side, basic observation shows that a cone's formula is 1/3 of the volume of a cylinder with the same base and height. Of course, I already revealed what the formula is, so the volume of a cylinder is \(\pi r^2 h\) , and the formula for a cone is \(\frac{1}{3} \pi r^2 h\)
.+2441 I have found a way to factor the original expression \(\left(x^2+y^2-z^2\right)^2-4x^2y^2\).
| \(\left(x^2+y^2-z^2\right)^2-4x^2y^2\\ \left(\textcolor{red}{x^2+y^2-z^2}\right)^2-(\textcolor{blue}{2xy})^2\) | Although this may not be immediately obvious, the original expression is a difference of squares. I rewrote the original expression to make this fact clearer. Remember that\(\textcolor{red}{a}^2-\textcolor{blue}{b}^2=(\textcolor{red}{a}+\textcolor{blue}{b})(\textcolor{red}{a}-\textcolor{blue}{b})\). Factor the expression as such. I have used colors to make this process easier for your eyes to digest. |
| \(\left(\textcolor{red}{x^2+y^2-z^2}+\textcolor{blue}{2xy}\right)\left(\textcolor{red}{x^2+y^2-z^2}-\textcolor{blue}{2xy}\right)\) | Let's do some rearranging here. |
| \(\left(\textcolor{green}{x^2+2xy+y^2}-z^2\right)\left(\textcolor{green}{x^2-2xy+y^2}-z^2\right)\) | Why am I highlighting this part? Well, both portions happened to be perfect-square trinomials. |
| \(\left([\textcolor{red}{x+y}]^2-\textcolor{blue}{z}^2\right)\left([\textcolor{red}{x-y}]^2-\textcolor{blue}{z}^2\right)\) | Do you notice something? Yes, it is another difference of squares. It is time to factor again! |
| \(\left([\textcolor{red}{x+y}]^2-\textcolor{blue}{z}^2\right)=(\textcolor{red}{x+y}+\textcolor{blue}{z})(\textcolor{red}{x+y}-\textcolor{blue}{z})\\ \left([\textcolor{red}{x-y}]^2-\textcolor{blue}{z}^2\right)=(\textcolor{red}{x-y}+\textcolor{blue}{z})(\textcolor{red}{x-y}-\textcolor{blue}{z}) \) | I have factored each one individually. The last step is to combine the factors into one expression. |
| \((\textcolor{red}{x+y}+\textcolor{blue}{z})(\textcolor{red}{x+y}-\textcolor{blue}{z})(\textcolor{red}{x-y}+\textcolor{blue}{z})(\textcolor{red}{x-y}-\textcolor{blue}{z})\) | This expression satisfies the original condition; the final expression is "the product of four polynomials of degree 1." |
+2441 1. Let's just pretend for the moment that \(c^3+4c>5c^2\) is an equation, so the equation would be \(c^3+4c=5c^2\). Let's solve this as if it were a regular equation.
| \(c^3+4c=5c^2\) | Subtract 5c^2 from both sides so that every term containing a "c" is on the left. |
| \(c^3-5c^2+4c=0\) | Factor out the GCF of the left-hand-side of the equation, c. |
| \(c(c^2-5c+4)=0\) | Now, let's factor some more. The trinomial is factorable. |
| \(c(c-4)(c-1)=0\) | Now, let's solve for all the possible values of "c." |
| \(c_1=0\\ c_2=1\\ c_3=4\) | |
Therefore, we have defined many possible ranges where the solutions of this inequality can lie. We will have to test all of them. The ranges are as follows:
\(c<0\\ 0 < c < 1\\ 0 < c < 4\\ c>4\)
Now, let's test values in these ranges.
For the range \(c<0\), let's test a point in this range, say -1, and see if it satisfies the original equation.
| \(c^3+4c>5c^2\) | Plug in -1 as a test point and see if it satisfies. |
| \((-1)^3+4*-1>5(-1)^2\) | If you use some logic here, you will realize that this test point does not satisfy the original inequality. (-1)^3 and 4*-1 are both negative values, so adding them together will result in a negative value, too. On the right hand side, we have a number squared, which will always be positive. Multiplying this result by 5 does not affect the signage. Therefore, the right hand side cannot be greater than the left, so it is a false statement. |
| \((-1)^3+4*-1\ngtr5(-1)^2\) | This means that any numbers in the range \(c<0\) are not solutions to the original inequality. |
Now, let's try the range \(0 . I will choose 1/2.
| \(c^3+4c>5c^2\) | Plug in the desired point. |
| \(\left(\frac{1}{2}\right)^3+4*\frac{1}{2}>5*\left(\frac{1}{2}\right)^2\) | This one does not appear to be as clear-cut, so we will have to perform more simplification. |
| \(\frac{1}{8}+2>\frac{5}{4}\) | Here, you can stop. 2 is already greater than 5/4. |
| This means that any real number in the range \(0 is a solution to the original inequality. |
Now, let's try the terciary range of \(1 . My test point, in this case, will be 2. Always pick a point that is easiest to plug in. Overcomplicating matters is only a burden to yourself.
| \(c^3+4c>5c^2\) | Evaluate if true when c=2. |
| \(2^3+4*2>5*2^2\) | This one appears to require some more iterations to determine. |
| \(8+8>20\) | I think it is obvious at this point that this is untrue. |
| \(8+8\ngtr20\) | As aforementioned, \(1 is not in the range. |
And finally, \(c>4\) is the last one. Let's do the next integer, c=5.
| \(5^3+5*2>5^1*5^2\) | You can use any tricks you desire to evaluate these inequalities quicker. Here is the one that I noticed straightaway. |
| \(5^3+5*2=5^3\) | Well, there is a 5^3 on both sides of the inequality. On the left, a 5*2 remains, so the left side has to larger. Notice how I refrained from actually evaluating this completely as to preserve precious time. |
| \(c>4\) is the range as well. |
Consolidating all this information, we can conclude that \(0 or \(c>4\). When you convert to inequality notation, you get \((0,1)\cup (4,+\infty)\)
.+2441 It appears as if this question has been edited since the replies of other answerers. As of now, the expression to simplify is \(\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
| \(\frac{2-\sqrt{3}}{2+\sqrt{3}}\) | Radicals in the denominator are considered to be improper; it is required that all radical terms are moved to the numerator. This can be achieved by utilizing the conjugate of the denominator: \(2-\sqrt{3}\). |
| \(\frac{2-\sqrt{3}}{2+\sqrt{3}}*\frac{2-\sqrt{3}}{2-\sqrt{3}}\) | To simplify the denominator, it is possible to use the algebraic rule that \((a+b)(a-b)=a^2-b^2\). Notice that this process does not change the inherent value of the expression since all I am doing is multiplying by 1, which has no effect. |
| \(\frac{\left(2-\sqrt{3}\right)^2}{2^2-\sqrt{3}^2}\) | By squaring the radical in the denominator, we are getting rid of a radical's existence from the denominator. |
| \(\frac{\left(2-\sqrt{3}\right)^2}{4-3}\) | Well, it is very clear at this point that the denominator simplifies to 1. Expand the numerator using the rule that \((a-b)^2=a^2-2ab+b^2\) |
| \(2^2-2*2*\sqrt{3}+\sqrt{3}^2\) | Simplify. |
| \(4-4\sqrt{3}+3\) | |
| \(7-4\sqrt{3}\) | This is what OfficialBubbleTanks got. |
