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Dimas [21]
2 years ago
6

What is this I need help don’t mind the room

Mathematics
1 answer:
beks73 [17]2 years ago
3 0
Here’s a picture of my work. The answer is b. Hope this helps :-)

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Using only the information derived in questions 2,3, and 4 which theorem or postulate can be used to prove abe=cde?
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using only the information derived in question 2, 3, and 4, which theorem

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What the simplified form of x- 1 +4+7x-3
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Collect like terms (x + 7x) + (-1 + 4 - 3)

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For the quadrilateral ABCD, E and F are midpoints of sides AD and BC respectively. AB = 15, BC = 29, CD = 34, AD = 11, and
uranmaximum [27]

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31

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7 0
2 years ago
Please help, thank you :)
wolverine [178]

Answer:

y^6

Step-by-step explanation:

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2 years ago
Question 8 Find the unit vector in the direction of (2,-3). Write your answer in component form. Do not approximate any numbers
slamgirl [31]

Answer:

The unit vector in component form is \hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}}  \right) or \hat{u} = \frac{2}{\sqrt{13}}\,i-\frac{3}{13}\,j.

Step-by-step explanation:

Let be \vec u = (2,-3), its unit vector is determined by following expression:

\hat {u} = \frac{\vec u}{\|\vec u \|}

Where \|\vec u \| is the norm of \vec u, which is found by Pythagorean Theorem:

\|\vec u\|=\sqrt{2^{2}+(-3)^{2}}

\|\vec u\| = \sqrt{13}

Then, the unit vector is:

\hat{u} = \frac{1}{\sqrt{13}} \cdot (2,-3)

\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}}  \right)

The unit vector in component form is \hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}}  \right) or \hat{u} = \frac{2}{\sqrt{13}}\,i-\frac{3}{13}\,j.

6 0
3 years ago
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