Can anyone help solve these math problems? Stuck!

[emitpeels]

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Gentlemen, it appears I have been stumped by some math questions while using a prep book. Math is not my strong point, so if you can solve them, please explain as simple as possible. For a few of the problems I have attached pictures. Please help!!

1) Determine the sets of points whose distances from (-2,2) and (3,-3) are in the ratios of 2:3.
Answer: Circle with center (-6,6) with radius 6 * Sqroot 2.

2) Which is equal to cos x if sec x = -5/3 and if pie/2 < x < 3pie/4?
Answer: -.6

3) Which represents the length of the side of a square that has the same numerical value for square units of area and units of perimeter?
Answer: 4

4) A person is offered a job with a first year salary of $30,000. Each succeeding year, she recieves a 10 percent cost of living increase. In addition, each year she is offered up to a $2,000 merit increase. Which of the following will be the max amount, in dollars, she earns during the third year?
Answer: $40,500.

5) A container holds 8 red and 4 white balls. Two balls are drawn in sequence - without replacement - from the container. Which represents the probability that exactly one of these balls is red?
Answer: 16/33
I got 14/33, maybe its an answer error?

6) What is the length of each side of a regular hexagon inscribed in a circle with a diameter of 12?
Answer: 6

7) An isosceles triangle has an 8 inch base and a height of 3 inches. Which is the measure of the angle included between the two equiv sides of the triangle?
Answer: 2 (arc tan 4/3)

If its isosceles, how can it have two different length sides?? How did they get this??

8) Two sides of an isoceles triangle have length 10. If the area is 48, what is the longest possible third side?
Answer: 16

Attached is two pictures of two additional questions I'm stuck on. If you can even answer just one of these I would be so grateful. I am confused... How can an isoceles triangle have two different sized legs if both legs should be the same length in an isos triangle? Help!!

 

[ICUrunning]

Are there no explanations in your book?

 

[gobinoob]

I can help with answering the nongeometric questions. Geometric questions are impractical to answer online, via forums. I would ask a teacher or something.

2. This question is about trigonometric identities. We are given, sec(x) = -5/3.

Recall the trigonometric identity: sec(x) = 1/cos(x)
So 1/cos(x) = -5/3. Then, we just flip both fractions to cos(x) = -3/5
The decimal form of -3/5 is -0.6
(You should write this all out on paper to see it better. I recommend using "top-down" fractions, not "left-right" as written)



3. For this problem, the fastest method is to set up and solve an algebraic equation:

Let the side length of our square = x
Then the perimeter (4x) will equal the area (x squared):

x^2 = 4x

x^2 – 4x = 0 (note, at this step, we should avoid this habit of cancelling x's on both sides, if we are looking for roots/solutions: "x^2 = 4x" cancelling to "x = 4"; although we get the right answer, we have "deleted" one of our possible solutions, x = 0, and in some cases, this solution can be part of the answer!)

x(x – 4) = 0
x = 0 and x = 4

(x = 0 is "discarded" because it makes no sense)
x = 4



4. For this question, I think we must assume that the 10% increase comes before the $2000 merit increase.
Recall that if we want to increase a number by 10%, the fastest method is to multiply it by 1.1… this is equivalent to, but faster than, adding 0.10 times the number to itself!

Year 1: 30,000
Year 2: 30,000 (1.1) = 33,000
Now, we add 33,000 + 2000 = 35,000
Year 3: 35,000 (1.1) = 38,500
Now, we add 38,500 + 2000 = 40,500



5. This is a binomial problem; I like to call these "a probability within a probability"
First, realize that there are multiple ways for us to select exactly one red ball. We could pick red first, white second (RW) or the opposite, (WR)
So let's find the chance that we pick RW:
(8/12)(4/11) = 8/33

But remember, there are two ways we can pick "exactly one red ball"! Hence, multiply by the number of ways, 2, to get 16/33
This is, in fact, a rather simple binomial problem. Here, our first probability was 8/33 and the second (the number of ways to arrange a red and a white) was just 2. Sometimes, this second probability will require a permutation or combination formula, as I will expand upon below:



Bonus problem: 10% of all dentists in the US are "bad". If we randomly select 5 dentists, what is the probability that exactly 3 will be "bad"?

The same steps as in #5.
First, we pick a random order. Let's go with BBBGG (where B = bad, G = good)
(1/10)(1/10)(1/10)(9/10)(9/10) = 0.00081

Second, we need to multiply this by the total number of possible orders in which we can arrange 3 B's and 2 G's. Here are three potential methods of thinking through this:

1. Imagine we want to choose 3 dentists out of a total of 5 dentists. How many ways can we choose 3 from 5? We utilize the "combination" formula (order does not matter, because BBBGG is the same as BBBGG (see, I rearranged two of the bad dentists there!))... which is 5 choose 3. So 5!/(3!2!) = 10. So we have 10 ways to choose 3 dentists out of 5.
2. Or we can imagine the dentists as just the letters B and G. This boils the question down to: how many ways can we arrange the word "BBBGG"... You might recall there is a formula for this as well, often known as the "TENNESSEE" formula! It is N!/(A!B!...) and in this case, the answer is identical: 5!(3!2!) = 10
3. Or we can write them all out. This is the method we selected for problem #5, because it was efficient for that problem. With 10 orders, this would be most inefficient for this problem!

Hence, multiply the two probabilities by one another: (0.00081)(10) = 0.0081
0.81%

Note: it is critically important to understand problems such as #5 and the bonus problem. You stand a very high chance of getting "a probability within a probability" problem on the test! Note that the bonus problem is about as hard as these get, however. You are not likely to be asked something like "what is the probability that at least 3 out of 10 selected dentists are bad". But if you wish, test yourself to see if you know how to solve that as well!

 

[emitpeels]

Are there no explanations in your book?

No explanations, only answers. Why would I post them if explanations were given? 😉

Gobinnoob - Thanks for the help! Your explanations made me realise the silly mistakes I have made. I like your explanation for the ball problem. I can always figure out how to initially set up those problems, but I can never tell when to multiply the answer by two, since there are two possibilities in the order of how you can pick the balls. Also - do you have any advice for permutation / combination problems? I still struggle with telling when order matters or not. Thank you for your help, it is greatly appreciated!

Also for number 2 - Wow I did not realize how easy that question is. If I get a question like that in the future, do I ignore the jibberish that comes after it, like the " if 90 < x <pie*4/3 ?

 

[emitpeels]

bump

 

[KyoPhan]

Here's solution for #7 attached to the post. Basically, you cut the triangle in half to get a right triangle. From here, you use a little trig to get your answer.

Isoceles triangle have 2 sides that are similar to each other, and 3rd side that is different.

 

[gobinoob]

No explanations, only answers. Why would I post them if explanations were given? 😉

Gobinnoob - Thanks for the help! Your explanations made me realise the silly mistakes I have made. I like your explanation for the ball problem. I can always figure out how to initially set up those problems, but I can never tell when to multiply the answer by two, since there are two possibilities in the order of how you can pick the balls. Also - do you have any advice for permutation / combination problems? I still struggle with telling when order matters or not. Thank you for your help, it is greatly appreciated!

Also for number 2 - Wow I did not realize how easy that question is. If I get a question like that in the future, do I ignore the jibberish that comes after it, like the " if 90 < x <pie*4/3 ?

Ok so I'm not the best person to ask about when order matters, as it can get quite complicated. Let me try my best to explain: order matters when the tangible outcome is different after reversing two items in the order. Here are some examples:

I have 5 friends and I want to pick 2 to go with me to the mall. Order doesn't matter here. Whether I bring Josh and Sam vs Sam and Josh, we're all going as a group. This is, in fact, the same reasoning for why order doesn't matter when we pick "3 bad dentists out of 5". However, once we introduce some element of ranking, order starts to matter. This can be as obvious as first and second place medals to be given to 2 out of 5 people (or perhaps titles like "president" and "vice president" to be given to 2 out of 5) or it can be as subtle as distinct positions in space, such as 2 musical chairs, and 5 people to sit in them (or a more common example is seats in a car; it makes a difference whether Josh is sitting in the front passenger seat or in the back seat). Here, the order is important!

Further, it can be even more subtle than that! Here are two problems I remember from practice:

1. A theater has 5 doors. What is the total number of paths we can take through the theater, entering through one door, and exiting through a different door? Notice this is just another "choose X out of Y total" problem, so we use either a permutation or combination; we are choosing 2 objects out of 5. In this case, order matters because the path that we take, if we go from door 1 to door 2, is different from (the opposite of) the path from door 2 to door 1. If we reverse the two doors, we travel a different path. Hence, when order matters, the tangible outcome will change if we reverse two items in our chosen selection.

2. We have 5 points on a grid, and we want to draw a line between two points. What is the total number of lines we can draw? This, again, is a "choosing" problem, but in this case, the tangible outcome is a line. The line will not change if drawn beginning at X and ending at Y vs beginning at Y and ending at X. Thus, order doesn't matter. Note just how similar this is to the previous problem; if we change the wording of the question, even a little bit, we can make order matter vs not matter.

Hope this helps!

 

[gobinoob]

Another thing that confused me all the time was how to remember which formula to use. The permutation (order matters) formula is n!/(n-r)! and the combination formula (order doesn't matter) is n!/((n-r)!r!).

Note the difference is just the presence or absence of the "r!". What I remember is: if order does matter, then we don't include the "r!" and if order doesn't matter, do include it. Or you could think it out rationally, by realizing that the inclusion of additional numbers in the denominator reduces the overall number of combinations (which should make sense because if order doesn't matter, then we have fewer distinct combinations in which to arrange our items) but thinking through all of that just confused me when all I wanted was quick access to the correct formula.

 

[neoba]

1) Recall the distance between two points (x1,y1) and (x2,y2)： D = sqrt((x1-x2)^2 + (y1-y2)^2) (sqrt: square root)
Assume the coordinate of any point that satisfies the requirement is (x,y)
The requirement "distances from (-2,2) and (3,-3) are in the ratios of 2:3" can be expressed as
sqrt[(x-(-2))^2+(y-2)^2]/sqrt[(x-3)^2+(y-(-3))^2]=2/3
or [(x^2+4x+4)+(y^2-4y+4)]/[(x^2-6x+9)+(y^2+6y+9)]=4/9
This eventually gives you
x^2+12x+y^2-12y=0, add 36+36 to both sides
(x^2+12x+36)+(y^2-12y+36)=72
(x+6)^2+(y-6)^2=72
(x-(-6))^2+(y-6)^2=(sqrt(72))^2=(6*sqrt(2))^2
This is exactly the "Circle with center (-6,6) with radius 6 * Sqroot 2"
There might be a simpler to solve the problem. But this is what I have at the moment.


6) Connect the center of the circle to all six corners of the hexagon. You will get six triangles, each of which has two sides = radius. Since all sides of the hexagon are of the same length (regular hexagon), it can be proved that any two triangles are congruent to each other (SSS), thus all six triangles are congruent. Therefore, the angle between any two adjacent radius is the same, 360/6 = 60 degrees, and all triangles are thus equilateral. Hence the side of the hexagon is equal to the radius of the circle = diameter/2 = 6.

8) Two sides of an isoceles triangle have length 10. If the area is 48, what is the longest possible third side?
Assume the unknown side's length is c, the angle opposite to it is C, and the other two sides of the same length, a.
Recall the area of a triangle: S=1/2*a*b*sin(C)
1/2 * a * a * sin(C) = 48, since a = 10, sin(C)=0.96
Since cos(C)^2+sin(C)^2=1, cos(C)=+/-sqrt(1-sin(C)^2)=0.28 (if C<90 degrees) or -0.28 (if 180>C>90)
Recall the law of cosines
c^2 = a^2 + a^2 - 2 * a * a * cos(C) =200-200*cos(C)=200+/-56=256 or 144
Therefore, the longest possible c would be sqrt(256)=16

The attached questions are hard to explain without actually drawing them (Is the first one 14/3, and the second one E?)