Answer:
AM ONLI GR 4 SORI HEHEhehe
Answer: I think it is b
Explanation:
Answer:
A - B = [ -1 3 ]
[ -6 2 ]
[ -4 -10 ]
Step-by-step explanation:
Since Matrix A and Matrix B have the same size, then we can simply perform the operation of subtraction on each position within the matrices to get the resulting subtracted matrix.
A = [ 4 7 ] B = [ 5 4 ]
[ -3 8 ] [ 3 6 ]
[ -5 -2 ] [ -1 8 ]
A - B = [ (4 - 5) (7 - 4) ]
[ (-3 - 3) (8 - 6) ]
[ (-5 - -1) (-2 - 8) ]
A - B = [ -1 3 ]
[ -6 2 ]
[ -4 -10 ]
Cheers.
9514 1404 393
Answer:
2√30 ∠-120°
Step-by-step explanation:
The modulus is ...
√((-√30)² +(-3√10)²) = √(30 +90) = √120 = 2√30
The argument is ...
arctan(-3√10/-√30) = arctan(√3) = -120° . . . . a 3rd-quadrant angle
The polar form of the number can be written as ...
(2√30)∠-120°
_____
<em>Additional comments</em>
Any of a number of other formats can be used, including ...
(2√30)cis(-120°)
(2√30; -120°)
(2√30; -2π/3)
2√30·e^(i4π/3)
Of course, the angle -120° (-2π/3 radians) is the same as 240° (4π/3 radians).
__
At least one app I use differentiates between (x, y) and (r; θ) by the use of a semicolon to separate the modulus and argument of polar form coordinates. I find that useful, as a pair of numbers (10.95, 4.19) by itself does not convey the fact that it represents polar coordinates. As you may have guessed, my personal preference is for the notation 10.95∠4.19. (The lack of a ° symbol indicates the angle is in radians.)
Answer:
lw + 
× π × 
⇒ Answer D is correct
Step-by-step explanation:
First, let us find the area of the semi-circle.
Area = × π × r²
<u>Given that,</u>
diameter of the semi-circle is ⇒ <em>l</em>
∴ radius ⇒ <em>l / 2</em>
<u>Let us find it now.</u>
Area = × π × r²
Area = × π ×
<u> </u>
Secondly, let us find the area of the rectangle.
Area = length × width
<u>Given that,</u>
length ⇒ <em>l</em>
width ⇒ w
<u>Let us find it now.</u>
Area = length × width
Area = l ×w
Area = lw
<u> </u>
And now let us <u>find the total area.</u>
Total area = Area of the rectangle + Area of the semi - circle
Tota area = lw + × π ×
