Here's another method to prove condition ii (\(\overline{AC}\perp\overline{DB}\)). I will utilize a twocolumn proof:
\(AD=CD\) \(AB=CB\)  Given 
\(\overline{AD}\cong\overline{CD}\) \(\overline{AB}\cong\overline{CB}\)  Definition of congruent segments 
Figure \(ABCD\) is a kite  Definition of a kite (If a quadrilateral has two unique pairs of sides that are congruent, then the figure is a kite) 
\(\overline{AC}\perp\overline{DB}\)  Property of a kite (Diagonals of a kite are perpendicular) 
This question has been answered at least 3 times on this forum. 6/2(1+2) evaluates to 9. Since this question has been asked multiple times on this forum, I have redirected you to 3 different explanations on how to get the answer:
1) https://web2.0calc.com/questions/6221#r2
Here's another method that works, too. Let's use the zeroproduct theorem. If haven't heard of the theorem, then you have probably used it before, unbeknown to you. With this theorem, we can make the following equation with the roots of \(\pm2\)
\((x+2)(x2)=0\)
The zeroproduct theorem states that at least one (or both) factors must be equal to zero for the lefthand side to equal 0. I have made an equation that contains both of the desired roots. Since we know that this equation has the desired roots and \(x^2+b10=0\) has the same desired roots, we can conclude that \((x+2)(x2)=x^2+b10\). Now, solve for b:
\((x+2)(x2)=x^2+b10\)  Expand the multiplication of both binomials by realizing that \((a+b)(ab)=a^2b^2\) 
\(x^24=x^2+b10\)  Subtract x^2 from both sides of the equation. 
\(4=b10\)  Add 10 to both sides. 
\(6=b\)  
This is simply another method of solving.
1)
When the question asks to solve for h in the equation of \(V=2\pi r^2h\), it means get h, alone, on one side of the equation. This only requires one step, in this case:
\(V=2\pi r^2h\)  Divide by \(2\pi r^2\) on both sides of the equation. 
\(h=\frac{V}{2\pi r^2}\)  We're done! h is by itself. 
The appropriate answer choice, therefore, is D
2) If the domain is only {2, 4, 7}, just plug them in the original equation of \(f(x)=x^25\).
\(f(2)=2^25=1\)  
\(f(4)=4^25=11\)  
\(f(7)=7^25=44\)  
Now, find the answer choice that has all three of these in its domain. That would be D again.
Maxwong has the correct idea, but I think missed the criterion that states that the triangle must be acute. With this restriction, there are not as many possibilities.
Have you ever heard of the Pythagorean Theorem? You probably have. But have you heard of the Pythagorean Inequalities Theorem? Maybe you have and maybe you haven't. I think my textbook summarizes the concept nicely and in a way that is fairly straightforward:
Source: http://abcpluspstudyguide.weebly.com/uploads/9/7/2/8/9728626/9817374.png?581
Of course, we seek triangles that only abide to acute triangles. We know 2 side lengths: 8 and 15. The remaining side I have labeled as s. Of course, there are 2 cases that are possible with s. Either it is the largest side or it is not. We will have to take both into account to solve this prolem:
For the first example, I will assume that s is the longest side, which corresponds to c in the picture above. Now, let's solve:
\(c^2  s is c. Whichever side you plug in for a and b does not matter.  
\(s^2<8^2+15^2\)  Simplify the right hand side of the inequality to solve for s  
\(s^2<64+225\)  
\(s^2<289\)  Take the square root of both sides.  
\(s<17\)  In an inequality, the absolute value has the following rule\(a . Now, let's apply it.  
 One inequality is already done. Divide both sides by 1 to get the other answer. Don't forget to flip the inequality sign!  
 We can clean up this solution by recognizing that we can represent the solutions as a compound inequality.  
\(17  
Now, let's solve for s when is not the longest side:
\(c^2  This time, when we substitute, b will be in either a or b.  
\(15^2  Simplify both sides.  
\(225  SUbtract 64 on both sides.  
\(161  Take the square root of both sides.  
\(\sqrt{161}<s\hspace{1mm}\text{or}\hspace{1mm}s>\sqrt{161}\)  The same logic applies for greater than symbols, too. In other words, \(a>b\rightarrow a>b\hspace{1mm}\text{and}\hspace{1mm} a>b\)  
 Just like above, divide by 1 on the right inequality.  
 These inequalities cannot be combined into a compound inequality, unfortunately.  
The triangle must adhere to both conditions of \(17 and ( \(s>\sqrt{161}\) or \(s<\sqrt{161}\) )
Let's think about this logicall before our head explodes, though! If s is less than 17 and greater than √161, that means the following compound inequality arises:
\(\sqrt{161}
We don't care about the other inequality as it contains negative numbers, and negative numbers are nonsensical in the context of this problem, so I have excluded them. We aren't done yet! We have to figure out the first integer that is greater that 161. Well, \(\sqrt{161}\) is in between 12 and 13, so 13 is the first integer that is allowable under our given conditions.
Therefore, the only possible integer solutions are 13, 14, 15, and 16.
In the second question, Cphill interpreted it correctly, but if you meant \(R(x)=\frac{3x3}{x^24}\) or \(R(x)=3x\frac{3}{x^24}\), the answer is different. Let's think about what we have to do to figure out the vertical asymptote.
In both cases, we have rational fractions. Of course, we can't have a denominator of 0. This means that if we plug in a value for x that results in a denominator that equals 0, then it is officially outside of the domain. Let's figure out when x^24=0.
\(x^24=0\)  You might notice that x^24 is a difference of 2 squares, but we don't need to take advantage of this, actually. This is because we have no bterm. Add4 to both sides. 
\(x^2=4\)  Take the square root of both sides. 
\(x=\pm2\)  
This means that the vertical asymptote is at \(x=\pm2\), which corresponds to the answer choice of C.
I have already answered this question. It was posted. Then, it disappeared into the stratosphere. That's strange.
In any case, usually, this expression would be in its simplified form already. However, in this case, the radicands (numbers inside of the radical) happen to be perfect squares, so this expression can be simplified further:
\(2\sqrt{4}+5\sqrt{9}\)  \(\sqrt{4}=2\hspace{1mm}\text{and}\hspace{1mm}\sqrt{9}=3\) 
\(2*2+5*3\)  Now, it is a matter of simplifying from here. 
\(4+15\)  
\(19\)  
No, the expression evaluates to 39:
\(3\left[\frac{308}{2}+2\right]\)  Do 308 first. 
\(3\left[\frac{22}{2}+2\right]\)  Do 22/2 next. Luckily, 2 divides into 22 evenly. 
\(3\left[11+2\right]\)  Do 11+2 after that. 
\(3[13]=3*13=39\)  
I'll help you simplfy this expression \(\left(\frac{1}{2}\right)^26\left(2\frac{2}{3}\right)\):
\(\left(\frac{1}{2}\right)^26\left(2\frac{2}{3}\right)\)  To abide to order of operations, I will start with the parentheses 2(2/3). 
\(\frac{2}{1}\frac{2}{3}\)  In order to subtract fractions, we must establish a common denominator. Currently, this isn't the case. To do that, we must figure out the LCM (lowest common multiple) of the denominators. In this case, one of the denominators is 1. That means that the LCM is whatever the other number is. The LCM is 3. Let's convert the fraction 
\(\frac{2}{1}*\frac{3}{3}=\frac{6}{3}\)  Note that I am really multiplying the fraction by 1, so I am not changing the value of the fraction. 
\(\frac{6}{3}\frac{2}{3}=\frac{4}{3}\)  Of course, when subtraction fractions, you only subtract the numerator. 
\(\left(\frac{1}{2}\right)^26*\frac{4}{3}\)  Yet again, to abide by the rules of the order of operations, I will now do the exponent. I will apply the rule that \(\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}\) 
\(\left(\frac{1}{2}\right)^2=\frac{1^2}{2^2}=\frac{1}{4}\)  I just applied the rule that I mentioned before. 
\(\frac{1}{4}6*\frac{4}{3}\)  Now, I will do the multiplication of 6*(4/3) 
\(\frac{6}{1}*\frac{4}{3}=\frac{24}{3}=8\)  This is the multiplication done now. Now, we can move on to the next step. 
\(\frac{1}{4}8\)  Yet again, manipulate 8 such that there are common denominators. 
\(\frac{8}{1}*\frac{4}{4}=\frac{32}{4}\)  
\(\frac{1}{4}\frac{32}{4}\)  Do the subtraction. 
\(\frac{31}{4}=7\frac{3}{4}\)  
Therefore, \(\left(\frac{1}{2}\right)^26\left(2\frac{2}{3}\right)=\frac{31}{4}=7\frac{3}{4}\)
Here is a picture of your diagram:
In this diagram, I have labeled all of the given info on it. Our goal is to figure out what \(\sin\angle G\) is. In righttriangle trigonometry, the sine function compares, from the angle of reference, the opposite angle to the hypotenuse. Let's do that:
\(\frac{\sin m\angle G}{1}=\frac{FH}{17}\)  Uh oh! We don't know what the length of the segment of FH. We can use Pythagorean theorem. If you are like me, I have memorized some Pythagorean triples (integer solutions to the Pythagorean theorem like 3,4, and 5), but I will solve for them anyway. 
Let's apply the Pythagorean Theorem to figure out the remaining side:
\(FH^2+15^2=17^2\)  Simplify both sides of the equation before continuing. 
\(FH^2+225=289\)  Subtract 225 from both sides. 
\(FH^2=64\)  Take the square root of both sides to undo the exponent of 2. 
\(FH=8\)  Of course, the absolute value splits the possible answers to its positive and negative form. 
\(FH=\pm8\)  Of course, in the context of geometry, a negative side length is nonsensical, so we will should reject the negative answer. 
\(FH=8\)  
Now that we know what FH is, we can plug it back into what we were originally solving for, \(\sin m\angle G\):
\(\sin m\angle G=\frac{8}{17}\)  Technically, I should stop here, as the question asked for the value of \(\sin m\angle G\), but I will solve for G just in case that is actually what you are solving for. To solve for the angle measure, you must use the inverse function. 
\(\sin^{1}\left(\frac{8}{17}\right)=m\angle G\)  Use a scientific calculator with trigonometric functions to approximate the measure of the angle. Since an angle is measured in degrees, the calculator should be in degree mode when the expression is inputted. 
\(m\angle G=\sin^{1}\left(\frac{8}{17}\right)\approx28.07^{\circ}\)  
We're done now!
We want to solve for b in the equation \(\frac{3}{4}b=2\). Let's do that!
\(\frac{3}{4}b=2\)  Multiply both sides of the equation by 4 to get rid of the denominator. 
\(3b=8\)  Divide both sides by 3 
\(b=\frac{8}{3}=2\frac{2}{3}=2.\overline{666}\)  
Let's think about this problem. There are 2 solutions to \(y!=1\). It is 0 and 1. Therefore, we need to set \(3x2\) equal to both 0 and 1:
\(3x2=0\)  Add 2 to both sides. 
\(3x=2\)  Divide by 3 on both sides. 
\(x=\frac{2}{3}\)  
Let's do the other one now!
\(3x2=1\)  Add 2 to both sides. 
\(3x=3\)  Divide by 3 on both sides. 
\(x=1\)  
Therefore, x=1 and 2/3
To do \(\left(\frac{1}{2}\frac{2}{5}\right)^2\), we will first have to convert both fractions into fractions that have common denominators. To do this, we must figure out the LCM (lowest common multiple) of both denominators in the expression. In this example, 2 and 5 are coprime, so simply multiply them together to figure out the LCM. The LCM is 2*%, or 10. Let's convert both fractions such that both have a denominator of 10. I will manipulate the fractions chronologically in which they appear in the expression:
\(\frac{1}{2}*\frac{5}{5}\)  Note that we are really multiplying the fraction by 1, so the actual value of the fraction will be unchanged. 
\(\frac{5}{10}\)  
And of course, the next fraction:
\(\frac{2}{5}*\frac{2}{2}\)  This accomplishes the exact same thing as abovegetting a common denominator. 
\(\frac{4}{10}\)  
Let's subtract the fractions, now:
\(\left(\frac{5}{10}\frac{4}{10}\right)^2\)  Subtracting the fraction involves subtracting the numerator and keeping the denominator unchanged. 
\(\left(\frac{1}{10}\right)^2\)  To square a fraction, use the following rule of \(\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}\). You are "distributing" the fraction to both the numerator and denominator. 
\(\frac{1^2}{10^2}\)  Simplify both the numerator and denominator. 
\(\frac{1}{100}\) 

Therefore, \(\left(\frac{1}{2}\frac{2}{5}\right)^2=\frac{1}{100}\)
I am coming to help! You want to know how to evaluate the expression \(\frac{4}{5}\frac{1}{6}\):
To subtract (or add) fractions, we have to create a common denominator. What does that mean? It means that the denominators of both fractions must be the same before we can actually do any subtraction. 4/5 and 1/6 are not fractions with common denominators, so we will have to manipulate the fraction such that both have a common denominator.
To do this, we must figure out the LCM (lowest common multiple) of both denominators in this expression. Normally, you could use a process to figure out the lowest common denominator, but we can actually use a shortcut here. If you are trying to find the LCM of two numbers that is 1 unit away from each other on a number line, then simply multiply both numbers together. This means that the LCM of 5 and 6 is 5*6, or 30:
This is the denominator that we want both fractions to have. Let's manipulate 4/5 first:
\(\frac{4}{5}*\frac{6}{6}\)  Notice that we are actually multiplying the fraction by 1, which does not change the actual value of the fraction. 
\(\frac{24}{30}\) 

Now, let's change the other fraction such that the denominator is 30:
\(\frac{1}{6}*\frac{5}{5}\)  Just like above, I am making the denominator to 30. 
\(\frac{5}{30}\) 
Now, let's subtracting the 2 fractions! It should be much easier now!
\(\frac{24}{30}\frac{5}{30}\)  When subtracting fractions, only subtract the numerator, but keep the denominator. 
\(\frac{19}{30}\)  19 and 30 are coprime, so the fraction cannot be simplified anymore. 
Therefore, \(\frac{4}{5}\frac{1}{6}=\frac{19}{30}\)