Guide :

# Math Summary

I need to work out this problem x2 - 14x + 74 = 0   But in step by step but written in summary as if i'm the tutor teaching and telling them step by step example...  suppose the problem was x(x+1)...then, a sample summary would be something like: take the x and multiply it out to each term in parenthesis...first we have x times x which is x^2 and we add this to x times 1 which is x to get x^2+x as our answer.

## Research, Knowledge and Information :

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## Suggested Questions And Answer :

### Math Summary

I need to work out this problem x^2 - 14x + 74 = 0 We're going to use a method called completing the square. Begin by subtracting 74 from both sides of the equation. x^2 - 14x + 74 - 74 = 0 - 74 x^2 - 14x = -74 Next, we divide the coefficient of x by the coefficient of x^2, then take 1/2 of that, and square it. We add that result to both sides of the equation. 14/1 = 14 14 * 1/2 = 7 7 * 7 = 49 x^2 - 14x + 49 = -74 + 49 x^2 - 14x + 49 = -25 We can factor the left side of the equation. (x - 7)(x - 7) = -25 The right side is a negative number. For this problem, it means there are no real roots, i.e., the graph of this equation does not cross the x-axis. There are imaginary roots, though. In order to obtain the square root of a negative number, we introduce i, which is the square root of -1. Using that, the answer to this problem becomes x - 7 = ±5i. Which means that x = 7 + 5i  and x = 7 - 5i.

### set theory

The Venn diagram shows the geometric regions associated with the described sets (math=maths). KEY: SO=science only, MO=math only; SM=science & math; PASS=those that passed; PSM=those that passed both subjects; PS=passed science only; PM=passed math only; NONE=those that neither took nor passed science or math. PSM=75 PS+PM+PSM=156 (NONE=0), PS+PM=156-75=81; PS=81-PM PS+PSM=2(PM+PSM) (twice as many passed science as passed math),  PS=2PM+PSM=2PM+75; Substitute for PS: 81-PM=2PM+75, 3PM=6, PM=2. So, two passed math only.

### You have 5 math books, 2 history books, and 6 science books. If the books are placed on a book shelf, what is the probability that each category is kept together?

Three are 6 ways of arranging the categories along the shelf: MHS, MSH, HMS, HSM, SMH, SHM, where M=math, H=history, S=science. We can take each arrangement of categories and work out the probability of each arrangement. Take MHS as an example. If M and H are organised to be kept together, then S automatically is kept together. There are 13 books, 5 being math books. So the probability of selecting a math book in the first position on the shelf is 5/13, leaving 12 books, 4 of which are math. The probability of selecting another math book is 4/12. So the probability of the first 5 books being all math is 5/13.4/12.3/11.2/10.1/9=1/1287. There are now 8 books left, 2 of which are history books. So the probability of selecting a history book next is 2/8, then 1/7 for the second history book. That's 2/8.1/7=1/28. The science books will all be together. Combine the math and history probabilities: 1/1287.1/28=1/36036. Now let's take MSH. Next to math we have science and the probabilities are 6/8.5/7.4/6.3/5.2/4.1/3=1/28. When we combine the probabilities we again get 1/36036. And so it goes on for the other permutations of the categories. So, adding the combined probabilities we get 6/36036=1/6006=0.01665%.

### In one semster 120 students did commerce. 38 did Ins 111,36 did Eng 111, 100 did math 111,26 did ins 111 and eng 111,22 did eng 111and math 111,26 did ins 111 and math111,18 did all 3

(a) The Venn diagram helps to solve this type of problem. Here's the key: A=Commerce students who didn't take any of the 3 subjects. B=Ins only students C=Math only D=Eng only E=Ins+Eng only F=Math+Eng only G=Ins+Math only H=All 3 Now some figures: A+B+C+D+E+F+G+H=120 B+E+G+H=38 D+E+F+H=36 C+F+G+H=100 E+H=26 F+H=22 G+H=26 H=18 So, E=8, F=4, G=8. C=100-(4+8+18)=70 D=36-(8+4+18)=6 B=38-(8+8+18)=4 A=120-(4+70+6+8+4+8+18)=2 (a) A=2 students took none of the 3 (b) C=70 students took math only (c) E=8 students took ins and eng only (d) B+C+D=80 students took only one course (e) E+F+G=20 students took only two courses

### Why do I have to take math?

Why do I have to take math? The answer the adults won't tell you is basically that you'll grow up stupid if you don't.  Everything else is nice words phrased in a positive, supportive manner (adult talk), but that's really it.  Or, in a competitive sense, the more the people around you get smarter, the dumber you get by comparison.  You have to keep up. The reason you have to take math is because math is a building block for a bunch of other stuff, both in school and in life.  You need math to understand pretty much everything.  If you can't count money, how do you know if you're getting the right change back when you buy something?  If you don't understand interest rates, is the car dealer giving you a good or bad deal?  If you don't understand averages and you're looking at swapping out a weapon in a video game, how do you know whether the new weapon is better or not?  If you can't multiply, how do you know how many calories you're getting from that packet of crisps? (only x calories per serving, but 8 servings per bag) The reason you have to take math is because it's a fundamental skill.  It's not some weird skill that isn't very useful now (how to read morse code, how to read semaphores, etc.).  It's a fundamental skill like walking or talking.  That's how important math is.

### What is Maths

?????????????? "maths" ???????????? aent no such anamal... the werd be MATH... MATH... MATH... MATH... MATH

### How many students studied math only?

To solve draw a large circle to represent 100 students. Inside the circle draw three smaller interlocking circles each representing a subject, represented by and labelled M, P, C. These circles produce 7 regions or areas caused by the overlaps and the region surrounding the three circles inside the larger circle is an 8th region. These are given letters a to h and represent the numbers of students in each region: a. M only b. P only c. C only d. M, P only e. M, C only f. All 3 g. P, C only h. None The combination of letters below represents addition of the quantities represented by the individual letters. abcdefgh=100; adef=40; bdfg=64; cefg=50; ef=20; df=24; c=8; f=10. (h=12, but this can be calculated as shown below.) abdefgh=92 and efg=42 (because c=8), so g=22 (because ef=20); ad=20; bg=40 (because df=24), so b=18 (because g=22); h=12 (because abdefgh=92, b=18, ad=20, efg=42); d=14 (because f=10), so a=6. There are a=6 students taking math only; b=18 students taking psychology only; g=22 taking psychology and chemistry only. (e=10 took math and chemistry only.)