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2 years ago

Please solve asap thanks

Mathematics

2 answers:

sasho [114]2 years ago

6 0

Answer:

A' (-3, 12)

B' (9, 6)

C' (-6, -6)

Step-by-step explanation:

A (-1 , 4 ) , K = 3 => K * each with coordinate of A => (-1 * 3, 4 * 3)

B ( 3 , 2 ) , K = 3 => K * each with coordinate of B => (3 * 3, 2 * 3)

C ( -2 , -2 ) , K = 3 => K * each with coordinate of C => (-2 * 3, -2 * 3)

tensa zangetsu [6.8K]2 years ago

5 0

I think..

A' =( -1,-4 )

B' = ( 3,-2 )

C'=( - 2, 2 )

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2 years ago

If m angle B=m angle D=41, find m angle C so that quadrilateral ABCD is a parallelogram

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ANSWER

139°

EXPLANATION

One of the interior angle properties of a parallelogram is that adjacent interior angles are supplementary.

This means that, the adjacent interior angles add up to 180°.

$m \angle \: B = m \angle \: D = 41$

Then

$m \angle \: C + m \angle \: D = 180 \degree$

$m \angle \: B + m \angle \:C= 180 \degree$

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3 years ago

How many pairs of parallel line does a triangle have

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3 years ago

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

4 0

3 years ago

Please solve asap thanks ​

Please solve asap thanks