Create a circle such that its center is point A and B is a point on the circle.

Mathematics

1 answer:

Andrej [43]3 years ago

8 0

Answer:

The center of a circle is the point in the circle which is equidistant to all the edges of thr circle. The point a is the center, while point b is an arbitrary point in the circle. Find attachment for the diagram.

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Which is the graph of f(x) = -(x + 3)(x + 1)?

ch4aika [34]

Answer:

Step-by-step explanation:

x intercepts are -3,-1

f(x)=-(x²+x+3x+3)=-(x²+4x)-3=-(x²+4x+4-4)-3=-(x+2)²+4-3

or f(x)=-(x+2)²+1

vertex is (-2,1)

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3 years ago

I NEED HELP! I WILL MARK BRANLIEST! I need help with 11 & 12

antiseptic1488 [7]

Answer:

The answer is 17 and no

Step-by-step explanation:

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3 years ago

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OLga [1]

Swapping rows alters the sign of the determinant:

$\begin{vmatrix} x & y & z \\ -8 & 2 & -12 \\ u & v & w \end{vmatrix} = - \begin{vmatrix} x & y & z \\ u & v & w \\ -8 & 2 & -12 \end{vmatrix}$

Multiplying a single row by a scalar scales the determinant by the same amount:

$\begin{vmatrix} x & y & z \\ u & v & w \\ -8 & 2 & -12 \end{vmatrix} = -2 \begin{vmatrix} x & y & z \\ u & v & w \\ 4 & -1 & 6 \end{vmatrix}$

Then

$\begin{vmatrix} x & y & z \\ -8 & 2 & -12 \\ u & v & w \end{vmatrix} = -(-2) \begin{vmatrix} x & y & z \\ u & v & w \\ 4 & -1 & 6 \end{vmatrix} = 2\times(-6) = \boxed{-12}$