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Mekhanik [1.2K]
3 years ago
6

What is the distance between two points located at (–6, 2) and (–6, 8) on a coordinate plane?"

Mathematics
1 answer:
ASHA 777 [7]3 years ago
3 0

Answer:

4 units

Step-by-step explanation:

X coordinate never changed so you just need to subtract 2 from 8 to see the unit difference.

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Find the bearing of the ship. Round to the nearest tenth of a degree if necessary.
adelina 88 [10]

9514 1404 393

Answer:

  117°

Step-by-step explanation:

We assume you want to report your bearing as an angle measured CW from North. For conventional angles greater than 90°, it will be ...

  450° -angle = 450° -333° = 117°

The bearing of the ship is 117°.

4 0
3 years ago
Help me pls I need this grade
Ivahew [28]

yes lol bigger hah trigger lol susush

5 0
3 years ago
Read 2 more answers
How Do I solve this I am quite confused
Kazeer [188]

Answer:

$11,714

Step-by-step explanation:

C(x) = 0.7x² - 462x + 87,944

this is a quadratic equation that opens up

the vertex will be the minimum

From the quadratic formula the x coordinate of the vertex is

x = -b/2a

x = 462/(2 * 0.7)

x = 330

Plug in

c(330) = 0.7(330²) - 462(330) + 87,944

c(330) = 11,714

$11,714

5 0
3 years ago
Whats 318,224,475 rounded too the nearest ten.​
aivan3 [116]
318,224,480

5 is in the ones place so u round up
8 0
3 years ago
Someone please let me know the answer ​
erastova [34]

Answer:

The answer is 30°

Step-by-step explanation:

when \:  \:   \sqrt{3}  \sin( \alpha )  -  \cos( \alpha )  = 0 \\  \sqrt{3}   \sin( \alpha )  =  \cos( \alpha )  \\ multiply \: by \:  \frac{1}{ \sqrt{3}  \cos( \alpha ) }  \\ then \\  \frac{ \sin( \alpha ) }{ \cos( \alpha ) }  =  \frac{1}{ \sqrt{3} }  \\  \tan( \alpha )  =  \frac{1}{ \sqrt{3} }  \\ then \:  \alpha  = 30

3 0
3 years ago
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