All categories

2 years ago

If p || r, then what is m∠G

Mathematics

2 answers:

grigory [225]2 years ago

8 0

The answer is 34!!!!!!

damaskus [11]2 years ago

7 0

There's not enough letters for me to show you how the answer was obtained. The answer is 34 however. Give the Brainly to the other responder. He answered first.

You might be interested in

Please help :)

romanna [79]

Answer:

Step-by-step explanation:

one calculator price =30/5=6

cost price of 6 calculator=6*6=36

6 0

2 years ago

Read 2 more answers

¿Cuál es el resultado de (-8x²-11x+14)+(6x²-8x+4)?

Neko [114]

(-8x²-11x+14)+(6x²-8x+4)

Combine términos iguales.

Este es su resultado:

-2x²-19x+18

7 0

3 years ago

Ruben swims several times per week in a lake near his home. Last summer, the average water temperature was 20 °C. This summer, t

docker41 [41]

16/20 = 0.8
0.8*100= 80
100-80= 20%

3 0

2 years ago

Solve five ninths plus four sixths.

Olin [163]

Answer:

11/9

or 1 and 2/9

or 1.22222222222222222222222 (2 goes on forever)

Step-by-step explanation:

3 0

3 years ago

show that the equation 3x2 + 3y2 + 6x-y= 0 represents a circle and find the centre and radius of the circle

Jlenok [28]

Answer:

Center of the circle: $(-1,\frac{1}{6})$

Radius of the circle: $\frac{\sqrt{37} }{6}$

Step-by-step explanation:

Let's start by dividing both sides of the equation by the factor "3" so we simplify our next step of completing squares for x and for y:

$3x^2+3y^2+6x-y=0\\x^2+y^2+2x-\frac{1}{3} y=0$

Now we work on completing the squares for the expression on x and for the expression on y separately, so we group together the terms in "x" and then the terms in "y":

$x^2+y^2+2x-\frac{1}{3} y=0\\( x^2+2x ) + (y^2-\frac{1}{3} y)=0$

Let's find what number we need to add to both sides of the equation to complete the square of the group on the variable "x":

$( x^2+2x ) = 0\\x^2+2x+1=1\\(x+1)^2=1$

So, we need to add "1" to both sides in order to complete the square in "x".

Now let's work on a similar fashion to find what number we need to add on both sides to complete the square for the group on y":

$(y^2-\frac{1}{3} y)=0\\y^2-\frac{1}{3} y+\frac{1}{36} =\frac{1}{36}\\(y-\frac{1}{6} )^2=\frac{1}{36}$

Therefore, we need to add " $\frac{1}{36}$ " to both sides to complete the square for the y-variable.

This means we need to add a total of $1 + \frac{1}{36} = \frac{37}{36}$ to both sides of the initial equation in order to complete the square for both variables:

$x^2+y^2+2x-\frac{1}{3} y=0\\x^2+y^2+2x-\frac{1}{3} y+\frac{37}{36} =\frac{37}{36} \\(x+1)^2+(y-\frac{1}{6} )^2=\frac{37}{36}$

Now recall that the right hand side of this expression for the equation of a circle contains the square of the circle's radius, based on the general form for the equation of a circle of center $(x_0,y_0)$ and radius R:

$(x-x_0)^2+(y-y_0)^2=R^2$

So our equation that can be written as:

$(x+1)^2+(y-\frac{1}{6} )^2=(\sqrt{\frac{37}{36}} )^2$

corresponds to a circle centered at $(-1,\frac{1}{6})$ , and with radius $\sqrt{\frac{37}{36}}=\frac{\sqrt{37} }{6}$

5 0

3 years ago