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bulgar [2K]
3 years ago
14

Hi can someone help me with the last two sets of shapes? Thank you so much! -Jenn

Mathematics
1 answer:
SOVA2 [1]3 years ago
8 0
You are correct there's at least one obtuse angle
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Option A. 3(x+8y)

Step-by-step explanation:

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Let v1=(3,5) and v2=(-4,7).
Nat2105 [25]
PART A

The given vectors are,

v_1 = \: < \: 3 , \: 5 \: >

v_2 = \: < \: - 4 , \: 7 \: >

The magnitude of the vector

v= \: < \: x , \: y \: > \:

is given by:

|v| = \sqrt{ {x}^{2} + {y}^{2} }

This implies that,

|v_1| = \sqrt{ {3}^{2} + {5}^{2} }

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|v_2| = \sqrt{ {( - 4)}^{2} + {7}^{2} }

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PART B

To find the unit vector in the direction of a given vector, we divide by the magnitude of that vector.

^{ - } _{v_1} = \: < \: \frac{3}{ \sqrt{34} } , \: \frac{5}{ \sqrt{34} } \: >

Rationalize the denominator.

^{ - } _{v_1} = \: < \: \frac{3\sqrt{34}}{ 34 } , \: \frac{5\sqrt{34}}{ 34 } \: >

Also,

^{ - } _{v_2} = \: < \: \frac{ - 4}{ \sqrt{65} } , \: \frac{7}{ \sqrt{65} } \: >

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PART C

The sketch of the given vectors as well as their unit vectors are shown in the attachment.

3 0
3 years ago
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