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3 years ago

How do I solve this plz help

Mathematics

1 answer:

svet-max [94.6K]3 years ago

4 0

Well it depends. If your radical is wrapped around the entire expression, then your answer would be 3xy²z²√10xz, but if your radical is ONLY wrapped around 90, then your answer would be 3√10x³y⁴z⁵ [radical wrapped ONLY around 10]. So, with the way this is written, although it is simple to figure this out, it is difficult to find the answer you are looking for.

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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

Help will make branliest

suter [353]

Answer:

I like that you watch wwe for one. but lets get started

the answer itself is m< 4=87

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Helppppppppppp please...

Ede4ka [16]

Order from left to right by row
2
3
80
1
-40
1/6
240

3 0

3 years ago

What is a + a^2 ?<br> Thank you<br> Don't write the steps, just the answer

Natalka [10]

Oh okie so cute lol lol is

7 0

3 years ago

Greg’s home is 1.8 miles from his school there is a music store halfway between school and his home today Greg wants to stop at

Darya [45]

I'm guessing the question is "how far does he have to walk to the store from where he is"

if he is walking home, then hmm
distance from store to school=1.8/2 since it is halfway, so 0.9 miles

he has walked 0.36 miles
so he needs to walk 0.90-0.36=0.54 miles

he needs to talk 0.54 miles to get to the store

8 0

3 years ago

How do I solve this ​plz help

How do I solve this plz help