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FinnZ [79.3K]
3 years ago
7

Given the magnitude of two vectors |⃗| = 4 and |⃗⃗| = 6, and |⃗ + ⃗⃗| = 8. Find the angle between the vectors ⃗ and ⃗⃗(when they

are placed tail to tail). = ∠(⃗, ⃗⃗)
Mathematics
1 answer:
Nutka1998 [239]3 years ago
3 0

Answer:

the angle between the vectors \overrightarrow{a} and \overrightarrow{b}  is  cos\phi  = \frac{1}{6}

Step-by-step explanation:

Given -

|\overrightarrow{a}| = 4 , |\overrightarrow{b}| = 6 ,  |\overrightarrow{a} + \overrightarrow{b} | = 8

If two vector \overrightarrow{a} and \overrightarrow{b} are inclined at an angle \phi

vector parallelogram method

|\overrightarrow{a} + \overrightarrow{b} |^{2} = |\overrightarrow{a}|^{2} + |\overrightarrow{b}|^{2} + 2 |\overrightarrow{a}||\overrightarrow{b}|cos\phi

64 = 16 + 36 + 48 cos\phi

cos\phi  = \frac{1}{6}

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2^3 times 2^3+= <br> what does this eqaul
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Answer:

36

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2x3 x 2x3 = 36

Sorry if this didnt help

3 0
3 years ago
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Adrian are 3 ani, iar tatăl lui are 34 de ani. Peste câți ani Adrian va fi de două ori mai tânăr decât tatăl său?
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Answer:

27 years

Step-by-step explanation:

Given that :

Adrian's age = 3

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Let the number of years = x

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3 years ago
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Anna71 [15]

Answer:

3,7/6

Step-by-step explanation:

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\sqrt{x-3} =0~gives~x=3\\or~\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} =0\\or~ \sqrt{x-1} +\sqrt{x+3} =\sqrt{2(2x-1)} \\squaring\\x-1+x+3+2\sqrt{x-1} \sqrt{x+3} =2(2x-1)\\2x+2+2\sqrt{(x-1)(x+3)} =4x-2\\2\sqrt{x^2-x+3x-3} =2x-4\\\sqrt{x^2+2x-3} =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=\frac{7}{6}

8 0
3 years ago
20
elena-s [515]

Answer:

12 in

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the scale factor of the enlarged frame is found by comparing the widths

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I hope this help you


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