Pls any1 who cn help find solution to ds problem shud do,tnx.
(1)a rectangle has one side 5cm longer dn d oda.d area of d rectngle is 300cm,fnd its perimeter.
(2)d sum of a number and its square is 30,find d number.
(3)d sum of ages of a mother and her daughter is 42.if d product of their ages is 243,three yrs ago.find their ages.

[b]
(1)a rectangle has one side 5cm longer dn d oda.d area of d rectngle is 300cm,fnd its perimeter.
Proper spelling next time please.
Let x = the other side since we know something about one side--the one that is 5cm longer. So we have two different lengths: x and x+5. x+5 is the side that is 5cm longer. Since we have the two different lengths, as a rectangle does, we can use the formula for area of a rectangle.
Area = length * height
Area = x * (x+5)
300 = x * (x+5) and solve for x.

When you do find x, use it to find the actual dimensions of the rectangle, remember there is x and there is (x+5).
Then enter this into the formula for the perimeter of a rectangle:
Perimeter = 2(length * height)

(2)d sum of a number and its square is 30,find d number.
Let x represent "a number". So, its square would be represented as x^2 (x to the power of 2). Now you can write a mathematical equation:
x + x^2 = 30, solve for x.

(3)d sum of ages of a mother and her daughter is 42.if d product of their ages is 243,three yrs ago.find their ages.
Let m represent the age of the mother now.
Let d represent the age of the daughter now.
So m + d =42

For the product three years ago, the age of the mother is (m-3) and the age of the daughter is (d-3). The mathematical equation is:
(m-3) * (d-3) = 243.
Here you can replace m with 42-d because that is what m equals from the equation above (m + d = 42) or you can replace d with 42 - m. You get that from the equation as well.
Solve for the sole variable, the letter in the equation.

Does this make sense? Any questions?
Also, what grade/class are you in and are you using a calculator or not? If so, what kind of calculator?
[/b]

Odunharry. I suggest you should find a way of doing these yourself, these questions must be seconday school questions 'cos they are elementary maths, which means you definately have teacher for it, so you can try taking it to him/her. Tunmi. Great job there!

@ d poster. Your questions need answer and that is what I would give. Some I wud giv HINTS.

1. If one side is 5cm is gr8ta dan d oda, it then means either the length or breadth is +5cm. Then we can say length is x+5cm and breadth is xcm. Area=L*B
A=LB
300=(X+5)X Expanding, we have
300=X^2+5X
Therefore,
X^2+5X-300=0. Then solve simultaneously to find X. After gettin X, u shud calculate knwin dat perimeter=2(L+B). N.B; UR L=x+5 and B =x wit our abov illustratn. U can as wel solve makin L=X and B=X+5

2. Let d no be p and d square wud b p^2. It then means that
p^2 + p=30.
P^2+p-30=0 solve simultaneously u wud get p=5 and -6 as ur answer

3. Let d mother age be x and d daughter be y. Then, x+y=42----(1) and xy=243----(2). Three yrs ago means that u make subractn frm each case by 3. Then solve simultaneously and ur answers wud fly out.
Try mor qtns on ur own!

tnx guys i ve tried 1 and 3 bt ws thnkin if its correct.

Whao! The question is vry elementry.bt the last poster mad a mistake. After suming. I.e (p+p^2=30) then u wl av (p^2+p-30=0) then solve the quodratic eqn. Using any method. The answer is -6 and +5.