Subect Help: Math - 09-01-2007, 03:51 PM
-Silver_Wolf_Kitty(Algebra, Geometry, Algebra II, Pre-Calculus)
-TheHayleyDoll(Algebra, Geometry, Algebra II, Others)
Post any questions about Math here. Question topics include, but are not limited to:
-Trigonometry & Calculus
Topic(please include the subtopic. ex--> Alegbra: Probability):
09-17-2009, 04:02 PM
Ok, so I'm not in honors, so this might sound easy, but I don't get it. We need to use the distributive property to rewrite these expressions....my teacher talks way to fast so I didn't get it all down...if you don't mind can you give steps? I have to do 6 of these, but I'll put one down.
-5(m+12) I'm probably wrong, but I got -5m+(-5)12
Not really understanding anything...
an example without () is....
18x-6 probably wrong again, but I got, 18(3x-1)
:P any help?
PS...he said no =
others: I might need help with
I put answers to all of these, but I wouldn't be surprised if I got them all wrong.
09-21-2009, 08:32 PM
Topic: Algebra -> formulas?
Question:I'm in geomerty this year, but right now we're doing a ton of Algebra 1 reviews, and I can't remeber how you answer this question:
y(squared) - 6y - 55
6t(squared) - t - 5
Can someone please help me with this? >_<
baby, you got me. ♥
09-21-2009, 10:06 PM
I need help with a question.It's HW due 2 days from now since tommorrow we have this thing so whatever.It's Pre-Algebra + Algebra.So it's kinda both.The question is:
6(to the power of 4)+3(t-6)=3t-7
I have to solve it.
Here's my work:
Is this wrong??
09-22-2009, 12:12 PM
iluvdolphin, you had the first one correct You are just supposed to distribute it throughout the parentheses. Same applies to the others with it.
As for the second, you made a minor mistake. If 18 is in front of the x, it is a multiple of a number. Notice how the last was 6, which is a multiple of 18. You divided each number inside by 6, but you kept the 18 for some reason even though its a multiple So, while you had the inside correct, the 18 should be 6. To check this, you would need to try to redistribute it and see if it equals the original problem. When doing problems like this, you should always see if the number without the letter could be divided into the one in front of the letter. If it can, you can reduce the numbers in it.
antimaie, do you remember how those formulas are factored? In the end, you are supposed to get something like (#x+/-#)(#x+/-#). I'll take your first problem and show you how.
y^2 - 6y - 55
So, looking at this problem, you know absolutely that the first letters of each parenthese does not have a number in front of it because there are none proceeding the ys, so you have part of your problem (y+/-#)(y+/-#).
Next, we look at the signs. Notice that they are -s, which means that there is a + in one and a - in the other parenthese because when you distribute the equation, a positive number times a negative number produces negative numbers. So, we know that it must now be (y+#)(y-#).
We are left now figuring out what numbers not only multiply to equal -55, but also will equal -6 when added together. You'll want to list out all of the multiples.
-11 x 5
11 x -5
-1 x 55
1 x -55
Those are all you can get with whole numbers. Next, you determine if any will equal -6. So, if you look below, I have added them together now.
-11 + 5 = -6
11 + -5 = 6
-1 + 55 = 54
1 + -55 = -54
So, the first pair is the only option, so your factored equation would be (y+5)(y-11).
This would change a bit will numbers in front of the "y" value. In your second problem, 6t^2 - t - 5, you begin differently, first figuring out what values will multiply together to get 6. That is 6 and 1, 2 and 3, the best option would be the 6 and 1 because you need to determine what numbers will add/subtract to equal -1(-6+5=-1). So, you begin with (6t+/-#)(1t+/-#). Next, you'll want to figure out what'll equal 5, which is simply 1 x 5. The tough part is then figuring out which one each number belongs in. If 6 is multiplied by 5, you will end up with -29, so you know now that 6 and 5 have to be in the same parenthese so they aren't multiplied together so your factoring is now at (6t+/-5)(1t+/-1). The final step is determining the signs. To get a -1, the 6 would have to be multipled with a negative number since it is larger, so you now have (6t+5)(1t-1). Just redistribute to check.
Grim, first off, with exponents(which is what I'm assuming you mean by "to the power of 4"), it is not the lower number times the exponent(what you did was 6 x 4 = 24), it is the lower number multipled by itself the exponent number's amount of times(in this case, 4, so it'd be 6 x 6 x 6 x6 = 1296). So, your equation would then be
1296 + 3(t-6) = 3t -7
not what you listed. Your second issue is when you distribute a number through parentheses(in this case 3(t-6)), you need to multiply everything incide the parentheses by the outside number, in this case positive three, so your problem would then be:
Strangely enough, I'm looking at that equation, and it wouldn't work because the ts would cancel each other out. To explain, you don't subtract the 3 from the t since they are being multipled together, so you would divide each side by three. Once that is done, you would want to isolate the t, but since they are equal to each other, they would cancel each other out(t-t=0). Are you sure that was written down right? That doesn't make sense and there is something seriously wrong with the equation since my only other thought is you are supposed to factor the left, but there is no variable with the exponent part.
09-23-2009, 10:02 AM
I think I might have now that you explained it.I think it was a 2 or 7.I have to go check.I'd feel so bad if it was and so stupid,too!Ugh.I just did.It was a 2 0.0 Ok then.Well since you explained it,I'll try my best to get it right.Thanx anywayz!
11-05-2009, 08:17 PM
Thank You for answering my last question :]
okk...here are some questions...i think i might have the answers right...but i suck at math so, idk. :P
here they are, and can you please show work so i can understand? THANK YOU
info you need to know:
Barry has 32 base ball cards: 14 Cincinnati Reds cards and 18 Cleveland Indians cards. He also has 25 football cards: 15 Cleveland Browns cards and 10 Cininnati Bengals cards.
1. If Barry picks one baseball card and one football card, what is the probability of picking and Indians card AND a Bengals card?
2. If Barry picks two football cards WITHOUT REPLACEMENT, what is the probability of picking a Bengals card first AND a Browns card second?
3. If two 8-sided number dice, each numbered 1 through 8, are rolled, what is the probability of getting a 3 on each die?
4. Erin's dentist has 8 blue toothbrushes and 12 green toothbrushes in a drawer. If the dentist takes two toothbrushes from the drawer without replacement, what is the probability that the first toothbrush will be blue AND the second toothbrush will be green?
5. Nate is thinking of a number 1 through 9. He is also thinking of a letter A through D. What is the probability that Hannah will correctly guess the number AND letter Nate is thinking of?
MY ANSWERS: [probably wrong]
11-05-2009, 10:31 PM
I'll explain this question by question so you can understand. These formulas will apply throughout and note that I included key words that will tell you when to use them:
probability=# of relevant outcomes/total # of outcomes
probability(and, with replacement)=relevant outcomes/total outcomes x relevant outcomes/total outcomes
probability(and, without replacement)=relevant outcomes/total outcomes x relevant outcomes(-1 if applicable)/total outcomes-1
probability(or)=relevant outcomes/total outcomes + relevant outcomes/total outcomes
1. First we will find the total number of outcomes, which is simply found by adding together all of the cards(32+25=57). Now, the relevant outcomes are 18(Indians) and 10(Bengals). To find the probability of choosing one of each(I'm assuming this is with replacement, so there isn't a missing card afterwards), you simply multiply. So it'd be the second formula from above:
18/57 x 10/57 = 180/3249
I'm figuring you know how to simpify, so I'm not going to explain that.
2. For this one, we are going to use our third formula. We already know from the given information that there are 25 football cards to chose from, so now we need to apply what we know about the types of cards here. The real difference is because we are not replacing a card, that means the number of cards are going to go down to 24 because there is one less possibility(though because we want different types of cards, the top number remains unaffected. If we wanted two bengals cards, it would also go down by 1). So we have:
10/25 x 15/24 = 150/600
3. First, we need to know the total number of outcomes for each roll, which is 8 as we were told. This won't change because we are using two different dies, each with 8 sides that cannot be taken away, so we are going to use formula two. Next, we need to know how many "3"s are on each die, which is 1 for each one. So our formula is:
1/8 x 1/8 = 1/64
4. As always, we determine the formula we are going to use. The keyword is "without replacement", so we will use formula three. Next, we want to find the total number of outcomes:
8 + 12 = 20
Now, we need to put in the relevant outcomes. Since our first pick is blue, the first number will be 8/20, and since the second pick is green, the second number will be 12/19(notice because the item of choice is different so we didn't subtract from the top number; we did for the bottom one though since one is gone from the total). So, our formula is:
8/20 x 12/19 = 96/380
5. This seems confusing, but it really isn't. What we are doing is the same this as above with the second formula, there just are different totals involved(9 and 4, respectively). Since these is only going to be one correct possibility for each, that means the relevant outcome for both is "1". So, our formula is:
1/9 x 1/4 = 1/36
You found this already, but it always helps to go over it again.
The key thing about probability is just understanding what goes where. Its confusing regardless, but you just have to keep going at it. If this doesn't make sense, please ask me to clarify confusing points. Probability is a tough thing to explain.
1 Equation Problem - 12-07-2009, 05:55 PM
Ok, so I should probably get this, but for some reason I don't...
If you could please show your work and show the value of q that would be great!!