a couple of maths questions?

Question

a) Factorise
9x squared -4

b) Solve the inequation:

5x + 2 / 11 is less than or equal to 2

c) Convert 180 kmh^ -1 to ms^ -1

d) If each corner of a square is joined to each other corner of the square, two diagonals will have been joined. To do the same for a hexagon, nine diagonals would have to be drawn. How many diagonals would have to be drawn on a regular polygon with 15 sides?

Answer 1

(a)
(3x+2)*(3x-2)

(b)
5x + (2/11) ≤ 2
5x ≤ 2-(2/11)
x ≤ (2-(2/11))/5
x ≤ 2.2

(c)
180km/hr * 1000m/km * 1hr/3600s = 50m/s

(d)
You can draw and you can count. *No* excuse for you not doing this one


Doug

Answer 2

d) ninety

A corner cannot draw a line to itself, nor to it's two neighbors, as a line to a neighbor would simply be a "side".
so, our first corner can draw x-3 lines.
The next corner can draw the same amount of lines as the first... it's neighbor who has already "went" does not interfere with this, because once again a line to that neighbor would be a side.
On the third corner, however, we notice that one line has already been drawn to us. Thus we subtract 4 now instead of three.
This, it turns out, continues evenly until the number we subtract is as large as x, which, in the case of a square, is on the third corner.
so for 15 we have this:
12+12+11+10+9+8+7+6+5+4+3+2+1 = 90

we can speed up the addition process by temporarily ignoring the first "12" and quick-summing the 12 to 1 countdown.
counting down from twelve to one involves saying 12 numbers, because we don't say zero. So we are going to say 12 numbers, but on average we are going to say exactly 6.5 every time. Or to put it another way, we are going to say 6 groups of two numbers each, where each group totals to 13... (13 divided by 2 is 6.5) The groups would be 12 + 1, 11 + 2, 10 + 3, 9 + 4, 8 + 5, and 7 + 6. So 6 times 13 is 78 plus our original 12 from the first corner is 90. That little trick works in a variety of situations.

Answer 3

a) 9 * 4^2 = 9 * 4 * 4 = 134

b) Note: "<=" to be read as less than or equal to
5x + 2 / 11 <= 2
You multiply everything with 11:
55x + 2 <= 22
55x <= 22 - 2
55x <= 20
x <= 20 / 55

c) 1km = 1000 m; 1h = 3600s
180 km / h = 180 * 1000 m / 3600 s = 50 m / s

d) Well, I could not solve the problem only from the data you gave.....sorry

But, I found out that for a 5 sides polygon....you need 5 diagonals, and for a 7 sides polygon.....14.

If you notice:

SIDES.................NR. of DIAGONAL
4..........................2
5..........................5=2+3
6..........................9=5+4
7........................14=9+5

So, my conclusion:
if you note
s= numbers of the sides of the polygon
d= numbers of diagonals needed

d= 2 + 3 + 4 + ....... + (s - 2)

In your case, s= 15, so

d= 2 + 3 + 4 + .... + 13 = 90

The answer: 90 diagonals would have to be drawn.

Answer 4

this is only the question b:

5x+2/11 take 2 = 5x/9

5x/9 divide by 5 = x/1.8

x is 1.8 which is less than or is equal to 2

Answer 5

Sori i can t ans ur ques coz i'm 12 n don t noe anything bout it

Answer 6

I think the first one is (3x+2)(3x-2)

Answer 7

a=387420485
b=less than
c=???
d=16
i think idk...