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lukranit [14]
3 years ago
6

SOMEONE PLEASE HELP

Mathematics
1 answer:
otez555 [7]3 years ago
8 0

Answer: c and e

Step-by-step explanation:

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Help me with this question
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Can someone please help with #16
ollegr [7]

Answer:

∠ WZX = 50°

XW is not an altitude.

Step-by-step explanation:

16. See the attached figure.

XW is the angle bisector of ∠ YXZ, hence, ∠ WXY = ∠ WXZ

Now, given that ∠ YXZ = 8x + 34 and ∠ WXY = 10x - 13

Hence, ∠ YXZ = 2 ∠ WXY

⇒ 8x + 34 = 2(10x - 13)

⇒ 8x + 34 = 20x - 26

⇒ 12x = 60

⇒ x = 5.

Hence, ∠ XZY = ∠ WZX = 10x = 50° (Answer)

Now, ∠ WXZ = ∠ WXY = 10x - 13 = 37°

Hence, from Δ WXZ,

∠ WZX + ∠ WXZ + ∠ XWZ = 180°

⇒ 50° + 37° + ∠ XWZ = 180°

⇒ ∠ XWZ = 93° ≠ 90°

Hence, XW is not an altitude. (Answer)

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3 years ago
Use Cramer’s Rule to solve system of equation.
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\left\{\begin{array}{ccc}5x+2y=4\\3x+4y+2z=6\\7x+3y+4z=29\end{array}\right\\\\A=\left[\begin{array}{ccc}5&2&0\\3&4&2\\7&3&4\end{array}\right]\\\\\det A=5\cdot4\cdot4+3\cdot3\cdot0+2\cdot2\cdot7-7\cdot4\cdot0-3\cdot2\cdot5-3\cdot2\cdot4=54\\W_x=\left[\begin{array}{ccc}4&2&0\\6&4&2\\29&3&4\end{array}\right]\\\det W_x=4\cdot4\cdot4+2\cdot2\cdot29+6\cdot3\cdot0-29\cdot4\cdot0-3\cdot2\cdot4-6\cdot2\cdot4=108

W_y=\left[\begin{array}{ccc}5&4&0\\3&6&2\\7&29&4\end{array}\right]\\\det W_y=5\cdot6\cdot4+3\cdot29\cdot0+4\cdot2\cdot7-7\cdot6\cdot0-29\cdot2\cdot5-3\cdot4\cdot4=-162\\W_z=\left[\begin{array}{ccc}5&2&4\\3&4&6\\7&3&29\end{array}\right]\\\det W_z=5\cdot4\cdot29+3\cdot3\cdot4+6\cdot2\cdot7-7\cdot4\cdot4-3\cdot6\cdot5-3\cdot2\cdot29=324\\\\x=\dfrac{\det W_x}{\det A}=\dfrac{108}{54}=2\\\\y=\dfrac{\det W_y}{\det A}=\dfrac{-62}{54}=-3\\\\z=\dfrac{\det W_z}{\det A}=\dfrac{324}{54}=6

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What is the factored form of each expression?<br><br> 20x + 35y
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The factored form of 20x+35y would be 5(4x+7y). If you use the distributive property of multiplication, you can see that the expression has not changed. 
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