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larisa86 [58]
3 years ago
14

A library building is in the shape of a rectangle. Its floor has a length of (3x + 5) meters and a width of (5x − 1) meters. The

expression below represents the area of the floor of the building in square meters:
(3x + 5)(5x − 1)

Which of the following simplified expressions represents the area of the floor of the library building in square meters?

28x − 5
15x2 − 5
15x2 + 28x − 5
15x2 + 22x − 5
Mathematics
2 answers:
malfutka [58]3 years ago
5 0
We can use the Front Outside Inside Last (FOIL) method to expand the double bracket

(3x+5)(5x-1)

Front ⇒(3x)(5x)=15 x^{2}
Outside ⇒(3x)(-1)=-3x
Inside ⇒(5)(5x)=25x
Last ⇒(5)(-1)=-5

Put the four terms together we have
15 x^{2} -3x+25x-5, then collect like terms
15 x^{2} +22x-5
lorasvet [3.4K]3 years ago
4 0

Answer:

15 x^{2} +22x-5

Is the answer:)

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olya-2409 [2.1K]

Answer:

6 x 2.5 = 15

8 x 2.5 = 20

L x W = 300 square units

Step-by-step explanation:

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8+n the answer is 45
liq [111]

Answer:

n=37

Step-by-step explanation:

45-8=37

6 0
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GREYUIT [131]
The Answer is (D. No solution)
6 0
3 years ago
The length of a rectangle is 2 meters more than its width. The area of the rectangle is 80 square meters. What is the length and
insens350 [35]

Answer:

Step-by-step explanation:

Formula

A = L * W

Givens

W = W

L = W + 2

Solution

Area = L*W

Area = (W+2)*W = 80            Remove the brackets.

Area = W^2 + 2W = 80         Subtract 80 from both sides.

Area = w^2+2W-80=80-80  Combine

Area = w^2 +2W-80 = 0        Factor.

Area = (w+10)(w - 8) = 0

W + 10 = 0 won't work

W = - 10 which isn't possible

W- 8 = 0

W = 8

L = 8 + 2 = 10

The answer looks like A

8 0
3 years ago
1.What are the zeros of the polynomial function?
Lorico [155]
Let's to the first example:

f(x) = x^2 + 9x + 20

Ussing the formula of basckara

a = 1
b = 9
c = 20

Delta = b^2 - 4ac

Delta = 9^2 - 4.(1).(20)

Delta = 81 - 80

Delta = 1

x = [ -b +/- √(Delta) ]/2a

Replacing the data:

x = [ -9 +/- √1 ]/2

x' = (-9 -1)/2 <=> - 5

Or

x" = (-9+1)/2 <=> - 4
_______________

Already the second example:

f(x) = x^2 -4x -60

Ussing the formula of basckara again

a = 1
b = -4
c = -60

Delta = b^2 -4ac

Delta = (-4)^2 -4.(1).(-60)

Delta = 16 + 240

Delta = 256

Then, following:

x = [ -b +/- √(Delta)]/2a

Replacing the information

x = [ -(-4) +/- √256 ]/2

x = [ 4 +/- 16]/2

x' = (4-16)/2 <=> -6

Or

x" = (4+16)/2 <=> 10
______________

Now we are going to the 3 example

x^2 + 24 = 14x

Isolating 14x , but changing the sinal positive to negative

x^2 - 14x + 24 = 0

Now we can to apply the formula of basckara

a = 1
b = -14
c = 24

Delta = b^2 -4ac

Delta = (-14)^2 -4.(1).(24)

Delta = 196 - 96

Delta = 100

Then we stayed with:

x = [ -b +/- √Delta ]/2a

x = [ -(-14) +/- √100 ]/2

We wiil have two possibilities

x' = ( 14 -10)/2 <=> 2

Or

x" = (14 +10)/2 <=> 12
________________


To the last example will be the same thing.

f(x) = x^2 - x -72

a = 1
b = -1
c = -72

Delta = b^2 -4ac

Delta = (-1)^2 -4(1).(-72)

Delta = 1 + 288

Delta = 289

Then we are going to stay:

x = [ -b +/- √Delta]/2a

x = [ -(-1) +/- √289]/2

x = ( 1 +/- 17)/2

We will have two roots

That's :

x = (1 - 17)/2 <=> -8

Or

x = (1+17)/2 <=> 9


Well, this would be your answers.


7 0
4 years ago
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