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nirvana33 [79]
3 years ago
9

X^2 + y^2 =-6x-14-6y

Mathematics
1 answer:
skad [1K]3 years ago
7 0

Answer:

x=−3+√(−y−1)(y+5)

Step-by-step explanation:

thats if your solving for x if not to math w    l   ay

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Find the roots of f(x)=2x^2+7x+3 by using the quadratic formula
Oksana_A [137]

Answer:

f(x) = 2 ( x + 7/4)2  - 25/8

Step-by-step explanation:

4 0
2 years ago
A right rectangular prism is packed with cubes of side length 1/2 inch. If the prism is packed with 5 cubes along the length, 3
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The answer is C 11 1/4 or 11.25 Cubic inches
6 0
3 years ago
Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle
Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

(x-h)^{2} +(y-k)^{2}=r^{2}

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

(x+3)^{2} +(y-4)^{2}=r^{2}

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

r=\sqrt{(2-4)^{2}+(-6+3)^{2}}

r=\sqrt{(-2)^{2}+(-3)^{2}}

r=\sqrt{13}\ units

The equation of the circle is equal to

(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}

(x+3)^{2} +(y-4)^{2}=13

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

(10+3)^{2} +(4-4)^{2}=13

(13)^{2} +(0)^{2}=13

169=13 -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

(x-1)^{2} +(y-3)^{2}=r^{2}

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

r=\sqrt{(6-3)^{2}+(2-1)^{2}}

r=\sqrt{(3)^{2}+(1)^{2}}

r=\sqrt{10}\ units

The equation of the circle is equal to

(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}

(x-1)^{2} +(y-3)^{2}=10

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

(11-1)^{2} +(5-3)^{2}=10

(10)^{2} +(2)^{2}=10

104=10 -----> is not true

therefore

The point is not on the circle

The statement is false

7 0
3 years ago
Hello fellow people please asnwer this will give brainliest if right!
elena-14-01-66 [18.8K]

Answer:

B

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
In the morning, Mary walked 2 1/4 miles
slamgirl [31]

Answer:

7 1/8 miles.

Step-by-step explanation:

6 0
3 years ago
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