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olya-2409 [2.1K]
3 years ago
7

I will give you brainliest!!!!! Use the elimination method to solve the system of equations 3x-y=8 3x+3y=12

Mathematics
2 answers:
azamat3 years ago
6 0

Answer:

x=3

y=1

(3,1)

Step-by-step explanation:

Alik [6]3 years ago
4 0

Answer: c. (3, 1)

Step-by-step explanation:

(3x - y = 8) + (3x + 3y = 12) = 6x + 2y = 20

Divide by 2

3x + y = 10

Plug in 3 for x

3(3) + y = 10

9 + y = 10

10 - 9 = 1

y = 1

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Help Please! I will mark brainliest!
Damm [24]

Answer:

17

Step-by-step explanation:

4 × 3² = 36

2 × 7 = 14

36 ÷ 12 = 3

3 ± 14 = 17

7 0
3 years ago
Need help please. will give brainliest. what is 3x+y
yaroslaw [1]

Answer:

3x+y

Step-by-step explanation:

It's already in simplest form, you can't do anything else to it.

8 0
3 years ago
Marc left his house to drive to work. As he heads down his street, his speed increases steadily until he sees the stop sign at t
Anna35 [415]
The third graph is the correct answer :)
7 0
4 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
Clare is paid $90 for 5 hours of work. At this rate, how many seconds does it take for her to earn 25 cent?
Alla [95]
50 secs

as $90÷5hrs= 18 $ per hrs

$18÷60min=$0.30 per min

$0.30÷60sec=$0.005 per sec

$0.25÷$0.005= 50sec to make $0.25

so 50sec is the answer
7 0
4 years ago
Read 2 more answers
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