The equation of a circle is x2+y2+10x−4y−20=0 . What is the radius of the circle?

5-4. Solve each of the following equations for the indicated variable. Use your Equation Mat if it is helpful. Write down each o

balu736 [363]

QUESTION 1

The given equation is :

$2(y - 3) = 4$
We want to solve for y. This means we have to isolate y on one side of the equation while all other constants are also on the other side of the equation,

We first divide both sides by 2 to obtain,

$\frac{2(y - 3)}{2} = \frac{4}{2}$

We cancel out the common factors to get,

$y - 3 = 2$

We now group like terms to get,

$y = 3 + 2$

We simplify to obtain,

$y = 5$

QUESTION 2

The given equation is

$2x + 5y = 10$
We want to solve for x. This means that we need to isolate x on one side of the equation, while all other variables and constant are also on the other side of the equation.

This implies that,

$2x= 10 - 5y$

We now divide through by 2, to obtain,

$\frac{2x}{2} = \frac{10}{2} - \frac{5y}{2}$

This will now give us,

$x = 5 - \frac{5}{2} y$

QUESTION 3

The given equation is
$6x + 3y = 4y + 11$

We want to solve for y. This means that we need to isolate y on one side of the equation, while all other variables and constants are also on the other side of the equation.

We first of all group all the y terms on one side of the equation to obtain,

$6x - 11 = 4y - 3y$

We simplify to get,

$6x - 11 = y$

This implies that,

$y = 6x - 11$

QUESTION 4

The given equation is,

$3(2x + 4) = 2 + 6x + 10$

We want to solve for x. This means that we need to isolate x on one side of the equation, while all other variables and constant are also on the other side of the equation.

Let us first expand the brackets to get,

$6x + 12 = 2 + 6x + 10$
This implies that

$6x + 12 = 6x + 12$

We group like terms to get,

$6x - 6x = 12 - 12$

We now simplify both sides to get,

$0x = 0$

We don't want x to vanish, so let us try to divide both sides by zero to get,

$x = \frac{0}{0}$

This is an indeterminate form, which implies that, x has infinitely many solutions. This means that all the real numbers are solution to the equation.

$\therefore \: x \in \: R$

QUESTION 5

The given equation is,

$y = - 3x + 6$

We want to solve for x, so we add the additive inverse of 6, which is -6 to both sides of the equation to get,

$y - 6 = - 3x$

We rewrite this to obtain,

$- 3x = y - 6$

We divide through by -3 to get,

$\frac{ - 3x}{ - 3} = \frac{y}{ - 3} - \frac{6}{ - 3}$

This will give us,

$x = - \frac{y}{3} + 2$

or

$x= 2 - \frac{1}{3} y$

5 0

3 years ago

2 1/4 cups = how many pints

Aleks [24]

Answer:

1 1/8 or 1.125 pints

explanation:

Hope this helps:)

6 0

3 years ago

Jeremy is cutting a board into pieces that are 1 2/3 feet long . The board is 14 1/4 feet long . How many 1 2/3 -feet long piece

Solnce55 [7]

Answer:

Total number of 1 2/3 feet long pieces that will be cut=8 pieces

Step-by-step explanation:

This can be expressed as;

Total length of the board=length per piece×number of pieces

where;

Total length of board=14 1/4 feet

Length per piece=1 2/3 feet

Number of pieces=n

Replacing;

14 1/4=1 2/3×n

(57/4)=(5/3)n

n=(57/4)×(3/5)=8.55

Total number of 1 2/3 feet long pieces that will be cut=8 pieces

7 0

3 years ago

Read 2 more answers

What is the reason for Statement 3 of the two-column proof?

Ivenika [448]

Refer to the attached image.

Given:

The measure of $\angle JMK= 52^\circ$ and $\angle KML= 38^\circ$ .

Also, Three rays ML, MK, and MJ share an endpoint M. Ray MK forms a bisector as shown in the attached image and the bisector divides angle JML into two parts.

To Prove: $\angle JML$ is a right angle.

Proof:

Statements Reasons

1. $m \angle JMK=52^\circ$ Given

2. $m \angle KML=38^\circ$ Given

3. $m \angle JMK+m \angle KML=m \angle JML$

The reason for statement 3 is Angle addition postulate. As angle JML is composed of 2 angles that is angle JMK and angle KML. So by adding the measures of angles JMK and KML, we will get the measure of angle JML which is referred as Angle addition postulate.

4. $52^\circ+38^\circ = m \angle JML$ Substitution property of equality