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CaHeK987 [17]
3 years ago
5

The length of a rectangular storage room is 3 feet longer than its width. if the area of the room is 40 square feet, find the wi

dth.
Mathematics
1 answer:
xenn [34]3 years ago
7 0

Answer:

Width of rectangular storage room= 5 feet.

Step-by-step explanation:

Let x be the width of storage room.

We have been given that the length of a rectangular storage room is 3 feet longer than its width. So the length of storage room will be x+3.

We are also given that the area of the room is 40 square feet.

Since the area of a rectangle is length times width.

\text{Area of rectangle}=\text{Length* Width}

Let us substitute our given values in area formula.

40=x*(x+3)

Upon distributing x we will get,

40=x^2+3x

x^2+3x-40=0

Now let us factor out our quadratic equation using splitting the middle term.

x^2+8x-5x-40=0

x(x+8)-5(x+8)=0

(x+8)(x-5)=0

x+8=0 or x-5=0

x=-8 or x=5

Since width can not be negative, therefore, the width of rectangle will be 5 feet.

Let us verify our answer.

Length of rectangular storage room is 3 feet longer than its width. So length will be 5+3=8.

Given: Area=40 square feet.

5*8=40.

Hence, width of rectangular storage room is 5 feet.

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Answer:

0.50b+1.25h=20

Step-by-step explanation:

Let b be the number of balloons and h be number of party hats.

We have been given that balloons cost $0.50 each, so cost of b balloons will be 0.50b.

We are also told that party hats each cost $1.25, so cost of h party hats will be 1.25h.

Further, Joe wants to spend exactly $20 on the party supplies. We can represent this information in an equation as:

0.50b+1.25h=20

Therefore, the equation 0.50b+1.25h=20 represents the number of balloons ( b) and the number of party hats ( h) that Joe can buy spending exactly $20.

5 0
3 years ago
Solve the following equation algebraically:<br> 3 x squared = 12
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X = 2 or -2

either way it’s still 12 , hope this helps !
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kari74 [83]
The answer is C. 16/55 because P(Q and R) = P (Q) . P (R) (for independent events Q and R) 

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5 0
3 years ago
*please help word problem* Adam and Barb had a used book sale over the weekend. Adam sold 4 hardcover books and ten paperbacks a
vekshin1

Let us assume charge for a hardcover book = $ h

And charge for a paperback = $ p.

Adam sold 4 hardcover books and 10 paperbacks and earned $64.

So, we can setup an equation by using above statement.

4h + 10p = 64        -------------equation(1)

Barb sold 5 hardcover books and 7 paperbacks and made $58.

So, we can setup another equation by using above statement.

5h + 7p = 58         -------------equation(2)

Let us solve equation(1) and eqation (2) by elimination method.

Multiplying first equation by -5 and second equation by 4 and adding, we get

-20h -50p = -320

20h +28p =  232

___________________

    -22p = -88

Dividing both sides by -22, we get

-22p/-22 = -88/-22

p=4.

Plugging p=4 in first equation.

4h + 10(4) = 64

4h +40 =64

Subtracting 40 from both sides, we get

4h = 24.

Dividing both sides by 4, we get

4h/4 = 24/4

h=6.

Therefore, they charge $6 for a hardcover and $4 for a paperback.

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