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ra1l [238]
3 years ago
14

Giving brainliest to whoever answers this.

Mathematics
1 answer:
hjlf3 years ago
8 0

Answer:

5,1

Step-by-step explanation:

7x + 4y = 39

2x + 4y = 14

since, the 4y same for both equation then we can eliminate the 4y by subtracting the two equation

7x + 4y = 39

2x + 4y = 14

---------------- -

5x + 0 = 25

5x = 25

x = 25/5 = 5

then we can input (subtitute) the value of x into the equation, I choose 2x + 4y = 14

2x + 4y = 14

2(5) + 4y = 14

10 + 4y = 14

4y = 14 - 10

4y = 4

y = 4/4 = 1

or if your teacher ask you to use elimination to find the value of x we can write like this

7x + 4y = 39

2x + 4y = 14

make the coeficient of x in both equation same

7x + 4y = 39 (x2)

2x + 4y = 14 (x7)

so

14x + 8y = 78

14x + 28y = 98

--------------------- -

0 - 20y = -20

-20y = -20

y = (-20)/(-20) = 1 (same answer right)

so the correct answe is

(5,1)

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find the equation of a circle which passes through the point (2,-2) and (3,4) and whose centre lies on the line x+y=2
Nadusha1986 [10]

Answer:

Equation of the circle

(x - 0.7)² + (y - 1.3)² = 12.58

Step-by-step explanation:

The formula for the equation of a circle is given as:

(x - a)² + (y - b)² = r²,

where(a, b) is the center of the circle and r = radius of the circle.

a) We are told in the question that the equation of the circle passes through point(2, -2)

Hence,

Substituting 2 for x and -2 for y in the equation of the circle.

(x - a)² + (y - b)² = r²

(2 - a)² +(-2 - b)² = r²

Expanding the bracket

(2 - a) (2 - a) + (-2 - b)(-2 - b) = r²

4 - 2a - 2a +a² +4 +2b +2b +b² = r²

4 - 4a + a² + 4 + 4b + b² = r²

a² + b² -4a + 4b + 4 + 4 = r²

a² + b² -4a + 4b + 8 = r²............Equation 1

We are also told that the equation of the circle also passes through point (3,4) also, where 3 = x and 4 = y

Hence,

Substituting 3 for x and 4 for y in the equation of the circle.

(x - a)² + (y - b)² = r²

(3 - a)² +(4 - b)² = r²

Expanding the bracket

(3 - a) (3 - a) + (4 - b)(4- b) = r²

9 - 3a - 3a +a² +16 -4b -4b +b² = r²

9 -6a + a² + 16 -8b + b² = r²

a² + b² -6a -8b + 9 + 16 = r²

a² + b² -6a -8b + 25 = r²..........Equation 2

The next step would be to subtract Equation 1 from Equation 2

a² + b² -4a + 4b + 8 - (a² + b² -6a -8b + 25) = r² - r²

a² + b² -4a + 4b + 8 - a² - b² +6a +8b - -25= r² - r²

Collecting like terms

a² - a² + b² - b² - 4a + 6a + 4b + 8b +8- 25 = 0

2a + 12b -17 = 0

2a + 12b = 17...........Equation 3

Step 2

We are going to have to find the values of a and b in other to get our equation of the circle.

Since the center of the circle(a, b) lies on x + y = 2

Therefore, we have

a + b = 2

a = 2 - b

2a + 12b = 17 ..........Equation 3

Substituting 2 - b for a in

2(2 - b) + 12b = 17

4 - 2b + 12b = 17

4 + 10b = 17

10b = 17 - 4

10b = 13

b = 13/10

b = 1.3

Substituting 1.3 for b in

a + b = 2

a + 1.3 = 2

a = 2 - 1.3

a = 0.7

hence, a = 0.7, b = 1.3

Step 3

We have to find the value of r using points (2, -2)

(x - a)² + (y - b)² = r²

Where x = 2 and y = -2

(-2 - 0.7)² + (-2 - 1.3)² = r²

(-2.7)² + (-3.3)² = r²

1.69 + 10.89 = r²

r² = 12.58

r = √12.58 = 3.55

Step 4

The formula for the equation of a circle is given as:

(x - a)² + (y - b)² = r²,

where(a, b) is the center of the circle and r = radius of the circle

a = 0.7

b = 1.3

r² = 12.58

Equation of the circle =

(x - 0.7)² + (y - 1.3)² = 12.58

7 0
3 years ago
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