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Mazyrski [523]
3 years ago
5

MATH PLEASE HELP TwT

Mathematics
1 answer:
kaheart [24]3 years ago
7 0

Answer:

The first and the third answer!

Step-by-step explanation:

The slope is NOT -6, it is -3/2 !!

The y intercept is 0

and the last one isn't right bc the slope and intercept is wrong!!!

I hope this helps, tho!! Have an awesome day :D

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What is the equation of a line that has a slope of 0 and a y-intercept of 4
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Answer:

y = 4

since you cant have y = 0x + 4, you drop the x and have a straight line (slope of 0) that runs all along the y point of 4.

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Add the following complex numbers: (5-3i)+(10+5i)
serious [3.7K]

Answer:

i = 7.5

Step-by-step explanation:

(5-3i)+(10+5i)

(5+10)+(5i-3i)

15+2i

15=-2i

15/2-i

7.5=i

6 0
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Below are the graphs of y=|x| and y=|x|-5. How are the functions related? The functions have the shape. The y-intercept of y=|x|
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same, 0, -5

Step-by-step explanation:

6 0
2 years ago
6.<br> Simplify the expression.<br> (-3)-4
Leto [7]

Answer:

-7

Step-by-step explanation:

  1. Re-write: -3 - 4
  2. -3 - 4 = -7

I hope this helps!

8 0
3 years ago
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3. At noon Lan is in his blue mini-van 300 km north of Makenna in her Bugatti. Lan heads south
LiRa [457]

Answer:

The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h

Step-by-step explanation:

At noon the location of Lan = 300 km north of Makenna

Lan's direction = South

Lan's speed = 60 km/h

Makenna's direction and speed = West at 75 km/h

The distance  Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km

The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km

The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km

Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.

By Pythagoras' theorem, we have;

s² = x² + y²

The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61

ds²/dt = dx²/dt + dy²/dt

2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt

2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900

ds/dt = 900/(2×30·√61) ≈ 1.92

The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h

5 0
3 years ago
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