Mid-Terms :(

**Rocher** · January 12th 2007, 04:59 PM

Okay, I just need help understanding and getting the answers for a few questions.

1) A woman's dress originally selling for $50.00 was marked down to yield 20% on cost. If the original profit was 33 1/3% on cost, what will be the new sales price.

2) Find the compound interest on $1000 for 9 months at 8% per annum, interest being calculated quarterly.

[And then I have the expand crap... I seriously don't understand this guys

Please pity me and help me out

]
You have to expand each of the following

3) a. (Y+3)(Y+4)
b. (2x-4)(x+8)
c. (7x-4)(3x-9)
d. (x-3)(x+3)
e. (x+3)(y-9)

4) Mr. Drake sold two books at $1.20 each. Based on the cost, the profit on one was 20% and the loss on th e other was 20% On the sale of the books, did he break even? Gain? Lost? Show how much he gained or lost.

5) There are 1600 pupils in a school. 3/8 of them are girls. 43% of the boys and 1/4 of the girls wear glasses. How many students in this school wear glasses?

6) If 12(exponent)x-4=1, What is the value of 3(exponent)x

7) If x is 3% of y and y is 7% of w, find x in terms of w.

Thanks!!!!!

Guys, I know this is alot

Sorry, but mid-terms are really hard and I really need to understand this stuff :P Kindly try and answer as many questions as possible by tommorow =)

**topsquark** · January 12th 2007, 05:09 PM

Originally Posted by Rocher

3) a. (Y+3)(Y+4)
b. (2x-4)(x+8)
c. (7x-4)(3x-9)
d. (x-3)(x+3)
e. (x+3)(y-9)

Do you know FOIL? (First, Outer, Inner, Last)

$(a + b)(c + d) = ac + ad + bc + bd$
ac is the product of the First terms in each factor
ad is the product of the Outer terms
bc is the product of the Inner terms
bd is the product of the Last terms.

a) $(y + 3)(y + 4) = y^2 + 4y + 3y + 12 = y^2 + 7y + 12$

b) $(2x - 4)(x + 8) = 2x^2 + 16x - 4x - 32 = 2x^2 + 12x - 32$

c) $(7x-4)(3x-9) = 7x^2 - 63x - 12x + 36 = 7x^2 - 75x + 36$

d) $(x-3)(x+3) = x^2 + 3x - 3x - 9 = x^2 - 9$

e) $(x+3)(y-9) = xy - 9x + 3y - 27$ and that's all we can do for this one.

-Dan

**Rocher** · January 12th 2007, 05:13 PM

Thank you. Actually I don't know FOIL. Teacher never taught me

However, can you tell me how you got those answers? Like for the first one, I have no idea how you got 7y+12. The 7 i understand, but the 12... How can you multiply and add them?

**topsquark** · January 12th 2007, 05:31 PM

Originally Posted by Rocher

Thank you. Actually I don't know FOIL. Teacher never taught me

However, can you tell me how you got those answers? Like for the first one, I have no idea how you got 7y+12. The 7 i understand, but the 12... How can you multiply and add them?

Originally Posted by topsquark

Do you know FOIL? (First, Outer, Inner, Last)

$(a + b)(c + d) = ac + ad + bc + bd$
ac is the product of the First terms in each factor
ad is the product of the Outer terms
bc is the product of the Inner terms
bd is the product of the Last terms.

a) $(y + 3)(y + 4) = y^2 + 4y + 3y + 12 = y^2 + 7y + 12$

First terms: $y \cdot y$
Outer terms: $y \cdot 4$
Inner terms: $3 \cdot y$
Last terms: $3 \cdot 4$

So
$(y + 3)(y + 4) = y^2 + 4y + 3y + 12$

$= y^2 + (4y + 3y) + 12 = y^2 + (7y) + 12$ <-- Adding up "like" terms

-Dan

NOTE: For a slightly more advanced explanation (which might help, might not) this is merely an application of the distributive law of multiplication over addition: a(b + c) = ab + ac.

$(y + 3)(y + 4)$

Let's momentarily define a = y + 3:
$(y + 3)(y + 4) = a(y + 4) = ay + 4a$

Now, put back in a = y + 3:
$(y + 3)(y + 4) = a(y + 4) = ay + a \cdot 4$ $= (y+3)y + (y+3)\cdot 4$

Now use the distributive law again (twice):
$(y+3)y = y^2 + 3y$
$(y + 3) \cdot 4 = 4y + 12$

(These two lines give use the terms in the FOIL expansion.)

Thus:
$(y + 3)(y + 4) = y^2 + 3y + 4y + 12$

**CaptainBlack** · January 12th 2007, 11:34 PM

Originally Posted by Rocher

3) a. (Y+3)(Y+4)
b. (2x-4)(x+8)
c. (7x-4)(3x-9)
d. (x-3)(x+3)
e. (x+3)(y-9)

Another way of expanding products like this is:

$(a+b)(c+d)=a(c+d) + b(c+d)=ac+ad \ +\ bc+bd$ .

I will only do one example with this, but here goes:

$(7x-4)(3x-9)=7x(3x-9) + (-4)(3x-9)=21x^2-63x + (-12)x + 36$

...... $=21x^2-63x -12x + 36$

Now collect together the multiples of $x$ into one term gives:

$(7x-4)(3x-9)=21x^2-75x + 36$

RonL