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navik [9.2K]
3 years ago
6

Write the standard equation for the circle center (–6, 7), r = 9.

Mathematics
2 answers:
HACTEHA [7]3 years ago
7 0

Answer:

Step-by-step explanation:

Given: The center of the circle is (6,7) and the radius is 9.

To find: The standard equation for the circle.

Solution: The standard equation for the circle with center (a,b) and radius ris given as:

(x-a)^2+(y-b)^2=r^2

Now, the center is (6,7) and the radius is 9, thus the equation of circle is:

(x-(-6))^2+(y-7)^2=(9)^2

(x+6)^2+(y-7)^2=(9)^2

which is the required equation of circle with center as (6,7) and the radius is 9.

Advocard [28]3 years ago
4 0
The standard equation for the circle centre can be written by using this formula: (x-h)² + (y -k)² = r² 

So, the equation in this case is: (x+6)² + (y-7)² = 81
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4 0
3 years ago
A 1.50 liter sample of dry air in a cylinder exerts a pressure of 3.00 atmospheres at a temperature of 25oC. Without change in t
cricket20 [7]
The answer is 4.5 l

To calculate this, we will use the Boyle's law using constant k, pressure P, and volume V:

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P1 = 3 atm

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3 atm * 1.5 l = 1 atm * V2

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V2 = 4.5 atm l / 1 atm

V2 = 4.5 l

5 0
3 years ago
Read 2 more answers
What is the next fraction in this sequence? Simplify your answer. 3/4, 1/4, 1/12, 1/36
antiseptic1488 [7]
Hey their,

3/4 
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1/12
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Hopethis helped


7 0
4 years ago
Use Cramer’s Rule to solve system of equation.
alisha [4.7K]

\left\{\begin{array}{ccc}5x+2y=4\\3x+4y+2z=6\\7x+3y+4z=29\end{array}\right\\\\A=\left[\begin{array}{ccc}5&2&0\\3&4&2\\7&3&4\end{array}\right]\\\\\det A=5\cdot4\cdot4+3\cdot3\cdot0+2\cdot2\cdot7-7\cdot4\cdot0-3\cdot2\cdot5-3\cdot2\cdot4=54\\W_x=\left[\begin{array}{ccc}4&2&0\\6&4&2\\29&3&4\end{array}\right]\\\det W_x=4\cdot4\cdot4+2\cdot2\cdot29+6\cdot3\cdot0-29\cdot4\cdot0-3\cdot2\cdot4-6\cdot2\cdot4=108

W_y=\left[\begin{array}{ccc}5&4&0\\3&6&2\\7&29&4\end{array}\right]\\\det W_y=5\cdot6\cdot4+3\cdot29\cdot0+4\cdot2\cdot7-7\cdot6\cdot0-29\cdot2\cdot5-3\cdot4\cdot4=-162\\W_z=\left[\begin{array}{ccc}5&2&4\\3&4&6\\7&3&29\end{array}\right]\\\det W_z=5\cdot4\cdot29+3\cdot3\cdot4+6\cdot2\cdot7-7\cdot4\cdot4-3\cdot6\cdot5-3\cdot2\cdot29=324\\\\x=\dfrac{\det W_x}{\det A}=\dfrac{108}{54}=2\\\\y=\dfrac{\det W_y}{\det A}=\dfrac{-62}{54}=-3\\\\z=\dfrac{\det W_z}{\det A}=\dfrac{324}{54}=6

5 0
3 years ago
Solve the system:<br> 2.5(x−3y)−3=−3x+0.5 3(x+6y)+4=9y+19
Aleksandr-060686 [28]

Answer:

The value of x and y that satisfy the equations is x = 2 and y = 1

Step-by-step explanation:

Given

2.5(x−3y)−3=−3x+0.5

3(x+6y)+4=9y+19

Required.

Find x and y

We start by opening all brackets

2.5(x−3y)−3=−3x+0.5 becomes

2.5x - 7.5y - 3 = -3x + 0.5

Collect like terms

2.5x + 3x - 7.5y = 3 + 0.5

5.5x - 7.5y = 3.5 ---- Equation 1

In similar vein, 3(x+6y)+4=9y+19 becomes

3x + 18y + 4 = 9y + 19

Collect like terms

3x + 18y - 9y = 19 - 4

3x + 9y = 15

Multiply through by ⅓

⅓ * 3x + ⅓ * 9y = ⅓ * 15

x + 3y = 5

Make x the subject of formula

x = 5 - 3y

Substitute 5 - 3y for x in equation 1

5.5(5 - 3y) - 7.5y = 3.5

27.5 - 16.5y - 7.5y = 3.5

27.5 - 24y = 3.5

Collect like terms

-24y = 3.5 - 27.5

-24y = -24

Divide through by - 24

y = 1

Recall that x = 5 - 3y.

Substitute 1 for y in this equation

x = 5 - 3(1)

x = 5 - 3

x = 2

Hence, x = 2 and y = 1

3 0
3 years ago
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