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mixer [17]
3 years ago
6

A circle is described by the equation x2 + y2 + 14x + 2y + 14 = 0. What are the coordinates for the center of the circle and the

length of the radius?
(-7, -1), 36 units


(7, 1), 36 units


(7, 1), 6 units


(-7, -1), 6 units
Mathematics
2 answers:
Alona [7]3 years ago
7 0
We are given the expression of the equation of a circle that is x2 + y2 + 14x + 2y + 14 = 0. Using ocmpleting the squares: 
x2 + y2 + 14x + 2y + 14 = 0(x+7)^2 + (y+1) ^2 = -14 + 49 + 1(x+7)^2 + (y+1) ^2 = 36 center thus is at (-7,-1) and the radius is equal ot square root of 36 equal to 6. 
irinina [24]3 years ago
3 0

Answer:

It's D people on this website dont know how to put the answer. smh

Step-by-step explanation:

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Each month cameron budgets $1040 for fixed expenses $980 for living expenses and $120 for annual expenses his annual net income
sineoko [7]
Let us first find how much he is spending every year, to do this let's find his monthly expense and multiply it by 12.


Every month Cameron spends 1040 + 980 + 120 = 2140

To find out how much he spends yearly, multiply the monthly value by 12,
2140 x 12 = 25680


This value is more than his net income so he clearly has a surplus, but to check we can subtract 129 from every month to get:

(1040 + 980 + 120 - 129) = 2011

2011 x 12 = 24132, which shows that his budget is saving 129 surplus every month. Choice B is correct.
5 0
3 years ago
Read 2 more answers
Determine whether the given vectors are orthogonal, parallel or neither. (a) u=[-3,9,6], v=[4,-12,-8,], (b) u=[1,-1,2] v=[2,-1,1
nevsk [136]

Answer:

a) u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168

Since the dot product is not equal to zero then the two vectors are not orthogonal.

|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}

|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}

cos \theta = \frac{uv}{|u| |v|}

\theta = cos^{-1} (\frac{uv}{|u| |v|})

If we replace we got:

\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

b) u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5

|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}

|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}

cos \theta = \frac{uv}{|u| |v|}

\theta = cos^{-1} (\frac{uv}{|u| |v|})

\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557

Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.

c) u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0

Since the dot product is equal to zero then the two vectors are orthogonal.

Step-by-step explanation:

For each case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

Part a

u=[-3,9,6], v=[4,-12,-8,]

The dot product on this case is:

u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}

|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}

And finally we can calculate the angle between the vectors like this:

cos \theta = \frac{uv}{|u| |v|}

And the angle is given by:

\theta = cos^{-1} (\frac{uv}{|u| |v|})

If we replace we got:

\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

Part b

u=[1,-1,2] v=[2,-1,1]

The dot product on this case is:

u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}

|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}

And finally we can calculate the angle between the vectors like this:

cos \theta = \frac{uv}{|u| |v|}

And the angle is given by:

\theta = cos^{-1} (\frac{uv}{|u| |v|})

If we replace we got:

\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557

Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.

Part c

u=[a,b,c] v=[-b,a,0]

The dot product on this case is:

u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0

Since the dot product is equal to zero then the two vectors are orthogonal.

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4 years ago
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egoroff_w [7]
Here to solve the equation we need to plug in the value y as given. -5x+2(7x)=9. -5x+14x=9. 9x=9. X=1.
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In a system of 2 equation with unknowns how many solutions are possible
vfiekz [6]
There are 2 solutions
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Write the decimal expansion for <br> .
Slav-nsk [51]

Answer:

decimal?????

Step-by-step explanation:

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