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3 years ago

What is the standard form of the equation for this circle?

Mathematics

1 answer:

avanturin [10]3 years ago

6 0

Answer:

(x + 1)² + (y - 10)² = 4

Step-by-step explanation:

Standard form of the equation of a circle is given by,

(x - h)² + (y - k)² = r²

where (h, k) is the center of the circle and 'r' is the radius.

In the given question coordinates of the center A are (-1, 10) and radius of the circle is 2 units.

By substituting these values in the formula,

(x + 1)² + (y - 10)² = 2²

(x + 1)² + (y - 10)² = 4

Therefore, standard form of the equation of the given circle is (x + 1)² + (y - 10)² = 4

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3 years ago

The school theater department made $2,142 on ticket sales for the three nights of their play. The department sold the same numbe

anyanavicka [17]

Answer:

102 tickets

Step-by-step explanation:

Let's say the total number of tickets sold is n.

7n = 2142

n = 2142 ÷ 7

n = 306

They sold 306 tickets in total.

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Therefore, they sold 102 tickets each night.

I hope this helps :)

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3 years ago

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3 years ago

Read 2 more answers

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago