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alexandr1967 [171]
3 years ago
14

PLEASE check these questions very carefully. Thank you very much.

Mathematics
1 answer:
Viktor [21]3 years ago
4 0

I can only see  #1,  #2, and  #3  in the picture.
I must say that I'm astonished.

Not only are all three of your answers correct,
but your work is neat, organized, and correct too.

You've attacked each problem in exactly the right way, the results
show it, and it's clear that you know what you're doing.

You rock !

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488, 460, 520, 544, 535<br> What is the range of the data?
n200080 [17]

Answer:

84

Step-by-step explanation:

To find the range, find the difference between the largest value and the smallest value.

544 - 460 = 84

7 0
3 years ago
Can anyone help me with this
liberstina [14]

Answer:

the answer is the won with ijkl

7 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
Someone plz help me please
Liula [17]

Answer:

c and d

Step-by-step explanation:

3 0
3 years ago
What values of b satisfy 3(2b+3)^2 = 36
Maslowich
\frac{-3 +/- 2 \sqrt{3} }{2}

Start by dividing both sides by 3. Then take the square root of both sides. Then subtract 3 from both sides and divide by 2.  
5 0
3 years ago
Read 2 more answers
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