ok smart guy who just asked for a hard question here you go.?

Question

1)The length of a rectangle is 3 feet more than twice the width. The perimeter is 36 feet. Find the length and width.

2)The perimeter of a rectangle is 72 cm. The width is 3 cm. less than half the length. Find the length and the width.

3)Salary +commision equals total earnings. Maria is paid $140 plus a commision of 20 percent on all sales over $300. What must her total sales be if she is to earn a total of $340?

Answer 1

1. W = 5 ft, L = 13 ft
2. W = 10 cm, L = 26 cm
3. $1300

Answer 2

1)The length of a rectangle is 3 feet more than twice the width. The perimeter is 36 feet. Find the length and width.

width = x

perimeter = 2(w+l) = 2(w + [3+2w]) = 36

3w + 3 = 36
3w = 33
w = 11, l= 25

2)The perimeter of a rectangle is 72 cm. The width is 3 cm. less than half the length. Find the length and the width.

72 = 2(L+w)

w = L/2-3

72 = 2(L+L/2-3) = 2L + L - 6
3L-6 = 72
3L=78 so L = 24, then w= 24/2-3 = 9

3)Salary +commision equals total earnings. Maria is paid $140 plus a commision of 20 percent on all sales over $300. What must her total sales be if she is to earn a total of $340?

total pay = 340
less slary of 140 , commision = 200.
well done maria. must have nice legs.

200 is 20 percent of what she sold, so that must have been 1000 sales over the 300, so total is 1300

Answer 3

The formula for Perimeter is:

P = 2L + 2W

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let

W = width

2W + 3 = Twice the width plus three feet more

36 = perimter in feet

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P = 2L + 2W

36 = 2(2W + 3) + 2W

36 = 4W + 6 + 2W

36 = 6W + 6

36 - 6 = 6W + 6 - 6

30 = 6W

30/6 = 6W/6

5 = W

The answer is W = 5

Insert the W value into the equation

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P = 2L + 2W

36 = 2[2(5) + 3] + 2(5)

36 = 2[10 + 3] + 10

36 = 2[13] + 10

36 = 26 + 10

36 = 36

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The width is 5 feet

The length is 13 feet

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Problem 2

P = 2L + 2W

- - - - - -

Let

lL = length

1/2L - 3 = Width

72 = Perimeter

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P = 2L + 2W

72 = 2L + 2(1/2L - 3)

72 = 2L + L - 6

72 = 3L - 6

72 + 6 = 3L - 6 + 6

78 = 3L

78/3 = 3L/3

26 = L

The answer is L = 26

Insert the L value into the equation

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Check

P = 2L + 2W

72 = 2(26) + 2[(1/2)(26) -3]

72 = 52 + 2[13 - 3]

72 = 52 + 2[10]

72 = 52 + 20

72 = 72

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The length is 26 cm

The width is 10 cm

- - - - - - -s-

Answer 4

Rich z is right, but Ill give you some pointers.

Call the width x feet. (feet, eh? Must be America!)
Then length is x + 3 feet.
Now add these together and double the result.
That's the perimeter so write that it equals 36.
Solve this equation and then write the answer as
length = ... ft; width = ... ft

2) Same idea but let x be length this time, then
width is x/2 - 3.

3) A bit harder.
Let sales be $x. Then "sales over $300" = $(x-300)
commission = 20*(x-300)/100
Now put this into the statement that
total earnings (i.e. sales + commission) = 340
and solve

Answer 5

1. Length = 13 feet Width = 5 feet

2. Length = 26cm Width = 10 cm

3. She needs to sell $1,000 dollars to get the $200 commision.

Answer 6

1. L = 2W+3; 2L+2W = 36; 4W + 6 + 2W = 36; W = 5; L = 13. The second problem is solved the same way.
3. The commission must be $200, which is 20% of 1000, so the total sales must be $1300.

Answer 7

1) L=13, W=5
2) L=26, W=10
3) $1,300

Answer 8

Try to go easy on her. Trust her when she says she's not going to cheat on you. My boyfriend is sometimes the way you are. He tends to be intimidated by my guy friends, but he is a bit more lenient than you are. If you love her that much, try not to cage her up and forbid her to see her friends or you run the risk of making her feel like you don't let her be herself, and possibly eventually want to break up with you. My boyfriend is also scared that I am going to cheat on him, and I tell him I'm not going to, and feels he has to be handsome/funny/smart, whatever, for me to stay faithful. To be honest, I'm not going to cheat on him because of any of those qualities. I am staying faithful because I LOVE HIM and for no other reason. If your girl does love you, you probably shouldn't worry and don't need to justify your qualities to keep her faithful.

Answer 9

Those are not hard questions. They are the ones from the first week in Beginning Algebra. To answer them just pay attention when they are explained in class. Otherwise you will be completely lost and flunking by the middle of the term.

Answer 10

I'm not even going to bother doing these questions. They're not difficult! What year are you in? 4?

I'll give you a hard question! simpliy this:

2(x+3)