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2 years ago

I don’t understand this i’m confused.

Mathematics

2 answers:

LenaWriter [7]2 years ago

5 0

1/256 would most likely be the answer

mihalych1998 [28]2 years ago

3 0

Answer:

(1/4) ⁴

= 1⁴/4⁴

= 1 / 256

Step-by-step explanation:

hope it helps you

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A point moves in the plane in the following sequence.

dem82 [27]

Answer:

Step-by-step explanation:

The answer is point D(-2,1).

1. If the point D was moved down 2 units, then its coordinates became (-2,-1).

2. If point (-2,-1) was reflected over the x-axis, then its coordinates became (-2,1).

3. If the point (-2,1) was moved 4 units to the right, then its coordinates became (2,1).

4. If point (2,1) was reflected over y-axis, its coordinates became (-2,1).

4 0

2 years ago

pls help with this immediately!!! (only answer if u know bc i cannot get this wrong so don’t try guessing. i’ll be giving braini

arsen [322]

Answer:

your answer should be $16.50 :)

Step-by-step explanation:

5 0

2 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:

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