please help me solve these problems every steps

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March 29th 2010, 08:45 PM
BeBeMala

please help me solve these problems every steps

a rectangle is three less than twice as long as it is wide. what are the length and the width if its perimeter is 42 inches?

a)15, 27 inches
b) 13,23 inches
c) 16, 8 inches
d) 13, 8 inches

if x is the first of three consecutive integers and the sum of the second and third is 208 then which of the following can be used to solve for x?

a) x+(x+2)+(x+4=208
b)2x+3=208
c)3x+3=208
d)2x+2=208

which is factor of: 5x^2 +27x+10

a) (5x+3)
b) (x+5)
c) (3x+5)
d) ( x+10)
March 29th 2010, 09:36 PM
harish21

Quote:

Originally Posted by BeBeMala

a rectangle is three less than twice as long as it is wide. what are the length and the width if its perimeter is 42 inches?

a)15, 27 inches
b) 13,23 inches
c) 16, 8 inches
d) 13, 8 inches

if x is the first of three consecutive integers and the sum of the second and third is 208 then which of the following can be used to solve for x?

a) x+(x+2)+(x+4=208
b)2x+3=208
c)3x+3=208
d)2x+2=208

which is factor of: 5x^2 +27x+10

a) (5x+3)
b) (x+5)
c) (3x+5)
d) ( x+10)

1. let l be the length of the rectangle and let w be the width.
then, by your question:
$l = 2w-3$

so the perimeler is given by : $2(l+w)= 2((2w-3)+w)$
since perimeter = 42, you have $2(2w-3+w) =42 \rightarrow 6w-6 = 42$ so, $w = 8$
then $l = 2w-3=2(8)-3 = 13$

so l=13 and w = 8

2. if x is the first integer, then $x+1$ and $x+2$ are the two consecutive integers after x. the question is saying that $(x+1)+(x+2)=208 \rightarrow 2x+3=208$

3. you can factorize $5{x^2} +27x+10$ as $5{x^2} + 25x+2x+27 = 5x(x+5)+2(x+5)=(5x+2)(x+5)$ so one of your factors is ...
March 29th 2010, 09:52 PM
BeBeMala

Quote:

Originally Posted by harish21

1. let l be the length of the rectangle and let w be the width.
then, by your question:
$l = 2w-3$

so the perimeler is given by : $2(l+w)= 2((2w-3)+w)$
since perimeter = 42, you have $2(2w-3+w) =42 \rightarrow 6w-6 = 42$ so, $w = 8$
then $l = 2w-3=2(8)-3 = 13$

so l=13 and w = 8

2. if x is the first integer, then $x+1$ and $x+2$ are the two consecutive integers after x. the question is saying that $(x+1)+(x+2)=208 \rightarrow 2x+3=208$

3. you can factorize $5{x^2} +27x+10$ as $5{x^2} + 25x+2x+27 = 5x(x+5)+2(x+5)=(5x+2)(x+5)$ so one of your factors is ...

Thank you, you made it very easy for me to understand...(Rofl)