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olganol [36]
3 years ago
12

HELP THIS IS THE LAST ONE I HOPE..​

Mathematics
2 answers:
saveliy_v [14]3 years ago
7 0
Your answer is either 11/3 or -11/3
Natalija [7]3 years ago
6 0

Answer:

FRACTION FORM:   x  =  33/2

MIXED NUMBER FORM:   x  =  16 and 1/2

DECIMAL FORM:   x  =  16.5

Step-by-step explanation:

(2x - 7) / 4      =      (x + 3) / 3

First, lets get rid of those fractions by multiplying both sides by 4 x 3...  (12)

6x - 21     =     4x + 12

<em />

<em>Next, we have to get all x's on one side, and all regular numbers on the other.</em>

 - Add 21 to both sides (to remove -21 from the left side)

6x - 21 + 21    =    4x + 12 + 21

   *simplify*.....

6x    =    4x + 33

 - Subtract 4x from both sides (to remove it from the right side)

6x - 4x    =    4x + 33 - 4x

    *simplify*.....

2x   =   33

 - Divide both sides by two (to get x by it's self)

FRACTION FORM:   x  =  33/2

MIXED NUMBER FORM:   x  =  16 and 1/2

DECIMAL FORM:   x  =  16.5

*Hope this helps!  If you have any questions, please feel free to ask.  Good luck!*

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0.125 = 125/1000

Step-by-step explanation:

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$23.8 million

Step-by-step explanation:

w = previous weekends earnings

16.2 = .68w

x = 16.2/0.68

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ALGEBRA NATION!!! NEED HELP :(
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6 0
3 years ago
Rafeeq bought a field in the form of a quadrilateral (ABCD)whose sides taken in order are respectively equal to 192m, 576m,228m,
Valentin [98]

Answer:

a. 85974 m²

b. 17,194,800 AED

c. 18,450 AED

Step-by-step explanation:

The sides of the quadrilateral are given as follows;

AB = 192 m

BC = 576 m

CD = 228 m

DA = 480 m

Length of a diagonal AC = 672 m

a. We note that the area of the quadrilateral consists of the area of the two triangles (ΔABC and ΔACD) formed on opposite sides of the diagonal

The semi-perimeter, s₁,  of ΔABC is found as follows;

s₁ = (AB + BC + AC)/2 = (192 + 576 + 672)/2 = 1440/2 = 720

The area, A₁, of ΔABC is given as follows;

Area\, of \, \Delta ABC = \sqrt{s_1\cdot (s_1 - AB)\cdot (s_1-BC)\cdot (s_1 - AC)}

Area\, of \, \Delta ABC = \sqrt{720 \times (720 - 192)\times  (720-576)\times  (720 - 672)}

Area\, of \, \Delta ABC = \sqrt{720 \times 528 \times  144 \times  48} = 6912·√(55) m²

Similarly, area, A₂, of ΔACD is given as follows;

Area\, of \, \Delta ACD= \sqrt{s_2\cdot (s_2 - AC)\cdot (s_2-CD)\cdot (s_2 - DA)}

The semi-perimeter, s₂,  of ΔABC is found as follows;

s₂ = (AC + CD + D)/2 = (672 + 228 + 480)/2 = 690 m

We therefore have;

Area\, of \, \Delta ACD = \sqrt{690 \times (690 - 672)\times  (690 -228)\times  (690 - 480)}

Area\, of \, \Delta ACD = \sqrt{690 \times 18\times  462\times  210} = \sqrt{1204988400} = 1260\cdot \sqrt{759} \ m^2

Therefore, the area of the quadrilateral ABCD = A₁ + A₂ = 6912×√(55) + 1260·√(759) = 85973.71 m² ≈ 85974 m² to the nearest meter square

b. Whereby the cost of 1 meter square land = 200 AED, we have;

Total cost of the land = 200 × 85974 = 17,194,800 AED

c. Whereby the cost of fencing 1 m = 12.50 AED, we have;

Total perimeter of the land = 576 + 192 + 480 + 228 = 1,476 m

The total cost of the fencing the land = 12.5 × 1476 = 18,450 AED

4 0
3 years ago
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