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galben [10]
3 years ago
15

PLEASE ANSWER FAST ASAP

Mathematics
1 answer:
AVprozaik [17]3 years ago
4 0

Answer:

Step-by-step explanation:

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23.78 to 1 decimal place
fomenos

Answer:

23.8

Step-by-step explanation:

4 0
3 years ago
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
he temperature dropped 12°F in 8 hours.If the final temperature was -7°F, what was the starting temperature?
anygoal [31]

Answer:

5°F

Step-by-step explanation:

The starting temperature is 5.

You get this answer when you add 12 to -7.

If the question is saying the temperature dropped 12°F every 8 hours then the answer would be 89°F because you multiply 12 and 8, then add 96 to -7.

Hope this helped a bit? :)

4 0
3 years ago
Which number line best shot the value of -6/5?
lozanna [386]

Answer:

B

Step-by-step explanation:

7 0
3 years ago
1) Mr. Charles bought dinner for his family.
Brilliant_brown [7]
The correct answer is A
8 0
3 years ago
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