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2 years ago

A scuba diver dives 243 feet into the ocean and then swims up 156

Mathematics

1 answer:

wariber [46]2 years ago

6 0

Answer:

Her final position would stop at -87.

Step-by-step explanation:

As she dove 243 feet into the ocean, that means she went down. Therefore she's currently at -243, however she swam up 156 feet ending her up at -87. If that was kinda complicated, then use -243+156 and you'll get -87.

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2 years ago

Question “If y =5, what is the value of the following expression <br><br> 7+2y²

yarga [219]

Answer:

Step-by-step explanation:

Hey there!

To find the solution, plug in 5 for y. When you do this you get, 7+2(5)². Square the 5 to get 25. And then you att 7+2(25). 25 x 2 is 50. 50 +7 is 57.

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2 years ago

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Show all work please!

tino4ka555 [31]

Length of DA: 90°
They are placed on a obtuse angle, which is 180° and DA are also placed on a right angle, so that means 180°-90°=90°

Length of AB: 180°
Their placed on a obtuse angle, therefore it will just be 180°

ADF=BDE because, both of them are right angles. This means their degrees will equal (90°=90°).

Hopefully this helps, sorry if I turn out to be wrong!

3 0

2 years ago

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

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

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

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

$\angle E=\arccos0=\dfrac{\pi}{2},\\ \\
\angle D=\arccos\letf(\sqrt{\dfrac{37}{83}}\right)\approx 0.27\pi,\\ \\
\angle F=\arccos\letf(\sqrt{\dfrac{46}{83}}\right)\approx 0.23\pi.$

3 0

2 years ago