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Thepotemich [5.8K]
2 years ago
7

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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This package of sliced cheese costs $2.97, How much would a package with 18 slices cost at the same price per slice
Mice21 [21]

Answer:

Cant you just type that in the calculator 18 times lol

Step-by-step explanation:

use a calculator

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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
Charra [1.4K]

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°.

If

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

Then

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

OR

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

\implies41 \degree +  m \angle \:C= 180 \degree

\implies m\angle \:C= 180 \degree - 41 \degree  = 139 \degree

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3 years ago
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3 years ago
How many pairs of parallel line does a triangle have​
sammy [17]

Answer:

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Step-by-step explanation:

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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
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