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3 years ago

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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Answer:

Option A. 3(x+8y)

Step-by-step explanation:

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3 years ago

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3 years ago

What equation is equivalent to 4,325,000,000 ?

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3 years ago

What is 1 2/9 divided by 2/3?

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2 years ago

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,

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

$|v_1| = \sqrt{9 + 25}$

$|v_1| = \sqrt{34}$

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

$|v_2| = \sqrt{ 16+ 49}$

$|v_2| = \sqrt{65}$

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,

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

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

PART C

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

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3 years ago