What is -2.75 as a fraction

Mrs. Culland is finding the center of a circle whose equation is x2 + y2 + 6x + 4y – 3 = 0 by completing the square. Her work is

andrew11 [14]

The correct answer is:

(x + 3)^2 + (y + 2)^2 = 4^2, so the center is (–3, –2)

Further explanation:

The given equation of circle is:

$x^2+y^2+6x+4y-3 = 0$

In order to complete the square the terms have to be combined first

$x^2+y^2+6x+4y = 3\\x^2+6x+y^2+4y=3\\(x^2+6x)+(y^2+4y) = 3$

9 and 4 will be added to both sides of the equation to complete the square

$(x^2+6x+9)+(y^2+4y+4) = 3+9+4\\(x^2+6x+9)+(y^2+4y+4) = 16$

The next step completes the work

$(x+3)^2+(y+2)^2 = (4)^2\\$

Comparing with the standard form of circle

$(x-h)^2 + (y-k)^2 = r^2\\x - h = x+3\\h=-3\\y-k = y+2\\k = -2$

As the centre is (h,k), the centre of given circle will be:

(-3,-2)

Hence the correct answer is:

(x + 3)^2 + (y + 2)^2 = 4^2, so the center is (–3, –2)

Keywords: Circle, Equation of Circle

Learn more about equations of circles at:

#LearnwithBrainly

4 0

3 years ago

Triangle RST is drawn inside rectangle RSNM. Point T is halfway between points M and N on the rectangle. The length of side RS i

Hunter-Best [27]

The question is not complete. The complete question is;

Triangle RST is drawn inside rectangle RSNM. Point T is halfway between points M and N on the rectangle. The length of side RS is 9 in. and the length of side RM is 8 in.

1) What is the area of triangle RST?

2)What is the ratio of area of triangle RST to area of triangle RMT?

3) What is the area of rectangle RSMN to area of triangle TSN?

Diagram is attached.

Answer:

1) 36 sq.in

2) 2:1

3) 4:1

Step-by-step explanation:

1) Area of triangle is; ½ × base × height.

For triangle RST, base = 9 in and height = 8 in

Thus, area of triangle RST = ½ × 9 × 8

= 36 sq.in

2)Area of triangle RMT = ½ × 8 × 9/2

= 18 sq.in

Ratio of area of RST to area of RMT = 36/18 = 2:1

3)Area of triangle TSN = ½ × 8 × 9/2 = 18 sq.in

Area of Rectangle RSNM = Length x width = 9 × 8 = 72 sq.in

Ratio of Area of Rectangle RSNM to Area of triangle TSN = 72/18 = 4:1

4 0

3 years ago

An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a d

anygoal [31]

Answer:

The probability of event A is $\frac{7}{11}$ .

The probability of event B is $\frac{6}{11}$ .

Step-by-step explanation:

The sample space of rolling a fair six-sided dice is as follows:

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Now consider the experiment of the computing the sum of the two rolls as follows:

X = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Number of total outcomes, N = 11.

The probability of an event E is the ratio of the number of favorable outcomes to the total number of outcomes.

$P(E)=\frac{n(E)}{N}$

The event A is defined as the sum is greater than 5.

The sample space of A is:

A = {6, 7, 8, 9, 10, 11, 12}

n (A) = 7

Compute the probability of event A as follows:

$P(A)=\frac{n(A)}{N}=\frac{7}{11}$

Thus, the probability of event A is $\frac{7}{11}$ .

The event B is defined as the sum is an even number.

The sample space of B is:

B = {2, 4, 6, 8, 10, 12}

n (B) = 6

Compute the probability of event B as follows:

$P(B)=\frac{n(B)}{N}=\frac{6}{11}$

Thus, the probability of event B is $\frac{6}{11}$ .

7 0

3 years ago

Sum of positive roots of the equation (x2 - 12x + 35) (x2 + 10x + 24) = 504 is

Digiron [165]

Answer:

(3)11

Step-by-step explanation:

We are given that

$(x^2-12x+35)(x^2+10x+24)=504$

We have to find the sum of positive roots of the equation.

$x^2(x^2+10x+24)-12x(x^2+10x+24)+35(x^2+10x+24)=504$

$x^4+10x^3+24x^2-12x^3-120x^2-288x+35x^2+350x+840=504$

$x^4-2x^3-61x^2+62x+840-504=0$

$x^4-2x^3-61x^2+62x+336=0$

Factor of 336

2,3,4,6,8,7,

Let x=2

$2^4-2^4-61(2^2)+62(2)+336\neq0$

x=2 is not the root of equation

x=-2

$(-2)^4-2(-2)^3-61(-2)^2+62(-2)+336=0$

Hence x=-2 is the root of equation.

x+2 is a factor of equation.

x=3

$3^4-2(3^3)-61(3^2)+62(3)+336=0$

Therefore, x=3 is the root of equation.

$(x+2)(x-3)(x^2-x-56)=0$

$(x+2)(x-3)(x^2-8x+7x-56)=0$

$(x+2)(x-3)(x(x-8)+7(x-8))=0$

$(x+2)(x-3)(x-8)(x+7)=0$

$x-8=0\implies x=8$

$x+7=0\implies x=-7$

Positive roots are 3 and 8

Sum of positive roots=3+8=11

Option (3) is true.

4 0

3 years ago

What is the value of j in the equation √ j + √ j + 14 = 3√ j + 10?

bulgar [2K]

J = 16 solve for j by simplifying both sides of the equation, then isolating the variable

3 0

3 years ago