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3 years ago

Which sign makes the statement true? 11 20 ? 61% > < Submit ASAP

Mathematics

1 answer:

Tcecarenko [31]3 years ago

3 0

11/20 < 61%

11/20 =.55
61% = .61
.55<.61

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Find the measures of the three angles, in radians, of the triangle with the given vertices: d(1,1,1), e(1,−5,2), and f(−2,2,7).

Oduvanchick [21]

Consider triangle DEF with vertices D(1,1,1), E(1,-5,2) and F(-2,2,7).

1. Find

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Then

$\cos \angle D=\dfrac{0\cdot (-3)+(-6)\cdot 1+1\cdot 6}{\sqrt{0^2+(-6)^2+1^2}\cdot \sqrt{(-3)^2+1^2+6^2}}=\dfrac{0}{\sqrt{37} \cdot \sqrt{46} }=0.$

2. Find

$\overrightarrow{ED}=(1-1,1-(-5),1-2)=(0,6,-1),\\ \\\overrightarrow{EF}=(-2-1,2-(-5),7-2)=(-3,7,5).$

Then

$\cos \angle E=\dfrac{0\cdot (-2)+6\cdot 7+(-1)\cdot 5}{\sqrt{0^2+6^2+(-1)^2}\cdot \sqrt{(-3)^2+7^2+5^2}}=\dfrac{37}{\sqrt{37} \cdot \sqrt{83} }=\sqrt{\dfrac{37}{83}}.$

3. Find

$\overrightarrow{FE}=(1-(-2),-5-2,2-7)=(3,-7,-5),\\ \\\overrightarrow{FD}=(1-(-2),1-2,1-7)=(3,-1,-6).$

Then

$\cos \angle F=\dfrac{3\cdot 3+(-7)\cdot (-1)+(-5)\cdot (-6)}{\sqrt{3^2+(-7)^2+(-5)^2}\cdot \sqrt{3^2+(-1)^2+(-6)^2}}=\dfrac{46}{\sqrt{83} \cdot \sqrt{46} }=\sqrt{\dfrac{46}{83}}.$

$\angle E=\arccos0=\dfrac{\pi}{2},\\ \\
\angle D=\arccos\letf(\sqrt{\dfrac{37}{83}}\right)\approx 0.27\pi,\\ \\
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3 years ago

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Troyanec [42]

Answer:

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3 years ago

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Alex787 [66]

Answer:

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4 years ago

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Roman55 [17]

Answer:

$f(x)= x-6\\g(x) = x^\frac{1}{2} (x+3)\\\left(x-6\right)\left(x^{\frac{1}{2}}\left(x+3\right)\right):\quad x^{\frac{5}{2}}-3x^{\frac{3}{2}}-18x^{\frac{1}{2}}\\x^2\sqrt{x}-3x\sqrt{x}-18x^{\frac{1}{2}}\\$

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3 years ago

3.4 is what percent of 4.25?<br><br> Enter your answer in the box.

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2 years ago