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3 years ago

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1 answer:

vekshin13 years ago

3 0

The answer is 6*square rooted*2

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For vectors u= (3,4) and v= (1,3) find CompuV and the angle between u and v.

vampirchik [111]

Answer:

The angle between the given vectors u and v is $\theta=cos^{-1}\left[\frac{3}{\sqrt{10}}\right]$

Step-by-step explanation:

Given vectors are $\overrightarrow{u}=(3,4)$ and $\overrightarrow{v}=(1,3)$

Now compute the dot product of u and v:

$\overrightarrow{u}.\overrightarrow{v}=(3,4).(1,3)$

$=(3)(1)+(4)(3)$

$=3+12$

$=15$

Now find the magnitude of u and v:

$|\overrightarrow{u}|=\sqrt{3^2+4^2}$

$=\sqrt{9+16}$

$=\sqrt{25}$

$=5$

$|\overrightarrow{u}|=5$

$|\overrightarrow{v}|=\sqrt{1^2+3^2}$

$=\sqrt{1+9}$

$=\sqrt{10}$

$|\overrightarrow{v}|=\sqrt{10}$

To find the angle between the given vectors

$\overrightarrow{u}.\overrightarrow{v}=|\overrightarrow{u}|\overrightarrow{v}|cos\theta$

$\theta=cos^{-1}\left[\frac{\overrightarrow{u}.\overrightarrow{v}}{|\overrightarrow{u}|\overrightarrow{v}|}\right]$

$=cos^{-1}\left[\frac{15}{5\times \sqrt{10}}\right]$

$=cos^{-1}\left[\frac{15}{5\times \sqrt{10}}\right]$

$\theta=cos^{-1}\left[\frac{3}{\sqrt{10}}\right]$

Therefore the angle between the vectors u and v is

$\theta=cos^{-1}\left[\frac{3}{\sqrt{10}}\right]$

3 0

3 years ago

4 + x/7 =2<br><br>Solve<br><br>a -14. b 10. c 12 d 42

Mnenie [13.5K]

Answer:

-14

Step-by-step explanation:

4 + x/7 =2

x/7= -2

x= -2 * 7

x=

-14

5 0

3 years ago

Please someone !!! Help me

jarptica [38.1K]

First, you would find a common denominator which is 12. You would then get 5/12 and 7/12 or 20/48 and 28/48. Using this you can see that their are 8 more girls than boys in the play. Hope you found this helpful!

3 0

3 years ago

How much of a radioactive kind of calcium will be left after 10 days if the half-life is 5 days

Sophie [7]

Answer: 16 g

Step-by-step explanation:

We can solve this with the Radioactive Half Life Formula:

$A=A_{o}.2^{\frac{-t}{h}}$

Where:

$A$ is the final amount of radioactive calcium

$A_{o}=64 g$ is the initial amount of radioactive calcium

$t=10 days$ is the time elapsed

$h=5 days$ is the half life of radioactive calcium

Solving with the given data:

$A=(64 g)2^{\frac{-10 days}{5 days}}$

$A=16 g$ This is the amount of radioactive calcium left

7 0

4 years ago

What is 33 1/3% of 24

Over [174]

It is 8 the answer your welco

7 0

4 years ago

Read 2 more answers