The price of the camera after this increase was 140% or 1.4
-1 beacuse its closer to the postives
<h3>The answer to your question is k (-3) = 21!</h3>
Here's how I got this answer:
<em><u>K(a) = 2a^2 - a </u></em>
<em><u>K(a) = 2a^2 - a K(-3) = 2(-3)^2 - (-3)</u></em>
<em><u>K(a) = 2a^2 - a K(-3) = 2(-3)^2 - (-3)K(-3) = 2(9) + 3</u></em>
<em><u>K(a) = 2a^2 - a K(-3) = 2(-3)^2 - (-3)K(-3) = 2(9) + 3K (-3) = 18 + 3</u></em>
<em><u>K(a) = 2a^2 - a K(-3) = 2(-3)^2 - (-3)K(-3) = 2(9) + 3K (-3) = 18 + 3K (-3) = 21</u></em>
I hope this helps!
Also sorry for the late answer, I just got the notification that you replied to my comment, I hope I came in time!
Answer:
Can't a computEr do thi$?
Step-by-step explanation:
32= Space
36= $
39= '
63=?
65-90 A-Z where A=65 and Z=90,
97-122 a-z where a=97 and z=122;
These are the necessary key to decoding this ASCII message in Base 10.
The Code
67 97 110 39 116 32 97 32 99 111 109 112 117 116 69 114 32 100 111 32 116 104 105 36 63
decoded is written as:
Can't a computEr do thi$?
Answer:
Step-by-step explanation:
<u>Find the m∠B:</u>
- m∠B = (360° - 310°) + 50° = 100°
a)
<u>Find AC using law of cosines:</u>
- AC = √(8² + 13² - 28*13*cos 100°) = 16.4 km (rounded)
b)
<u>Find m∠A using law of sines:</u>
- 16.4/ sin 100 = 13 / sin A
- m∠A = arcsin (13 sin 100 deg / 16.4) = 51° (rounded)
<u>The bearing of C from A:</u>
c)
<u>Let the required distance is x:</u>
- x/ BC = sin 50°
- x = BC sin 50°
- x = 13 sin 50°
- x ≈ 10 km (rounded)