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Nuetrik [128]
3 years ago
13

Pls help with math ASAP

Mathematics
1 answer:
9966 [12]3 years ago
7 0
I think 6,1 and the upper point is -6,11
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Calculate the mean 12,2,8,6,57
Rzqust [24]
12+2+8+6+57=85

85/5= 17

mean=17
6 0
3 years ago
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What is the volume of the prism?
anygoal [31]

Answer:

78.75 cm cubed

Step-by-step explanation:

To find the volume of the prism, use the formula V = l*w*h where l = 4.5, w = 5 and h = 3.5.

So the volume is V = 4.5*5*3.5 = 78.75

7 0
4 years ago
Hi can you please help me​
astra-53 [7]

Answer:

<h2>Area = 90 cm²</h2><h2>Perimeter = 53 cm</h2>

Step-by-step explanation:

area and perimeter of the given diagram

<u>area = 1/2 * base * height</u>

where base = 15 cm

        height = 12 cm

plugin values into the formula

area = 1/2 * 15 * 12

area = 90 cm²

<u>Perimeter = add up all sides (AB + AC + BC)</u>

where side BC = √(5² + 12²)

                   BC = 13 cm

Perimeter = 25cm + 15 cm + 13 cm

Perimeter = 53 cm

6 0
3 years ago
NEED ASAP
frez [133]

Answer:

3 over 12

Step-by-step explanation:

3 over 12

8 0
3 years ago
Read 2 more answers
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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