Answer:
Answer b of the first one
answer of q9
Step-by-step explanation:
(x+iy)(2−3i)=4+i
2x−(3x)i+(2y)i−3yi
2
=4+i
Real
2x+3y
+
Imaginary
(2y−3x)
i=4+i
Comparing the real & imaginary parts,
2x+3y=4--------------------------(1)
2y−3x=1----------------------------(2)
Solving eq(1) & eq(2),
4x+6y=8
−9x+6y=3
13x=5⇒x=
13
5
y=
13
14
∴(x,y)=(
13
5
,
13
14
)
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SIMILAR QUESTIONS
star-struck
If e
x+iy
=α+iβ, then x+iy is called logarithm of α+iβ to the base e.∴log
e
(x+iy)=log
e
(re
iθ
) =log
e
r+iθ where r is modulus value of x+iy & θ be the argument of x+iy If i
α+iβ)
=α+iβ, then α
2
+β
2
equals
Hard
View solution
>
The modulus of (1 + i) (1 + 2i) (1 + 3i) is equal to
Answer:
put the numbers in order from least to greatest and then cross them out til you get to the center number
let's recall that in a Kite the diagonals meet each other at 90° angles, Check the picture below, so we're looking for the equation of a line that's perpendicular to BD and that passes through (-1 , 3).
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of BD


so we're really looking for the equation of a line whose slope is -1/3 and passes through point A
![(\stackrel{x_1}{-1}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{-\cfrac{1}{3}}[x-\stackrel{x_1}{(-1)}]\implies y-3=-\cfrac{1}{3}(x+1) \\\\\\ y-3=-\cfrac{1}{3}x-\cfrac{1}{3}\implies y=-\cfrac{1}{3}x-\cfrac{1}{3}+3\implies y=-\cfrac{1}{3}x+\cfrac{8}{3}](https://tex.z-dn.net/?f=%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B3%7D%29%5Cqquad%20%5Cqquad%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20-%5Ccfrac%7B1%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-%5Cstackrel%7By_1%7D%7B3%7D%3D%5Cstackrel%7Bm%7D%7B-%5Ccfrac%7B1%7D%7B3%7D%7D%5Bx-%5Cstackrel%7Bx_1%7D%7B%28-1%29%7D%5D%5Cimplies%20y-3%3D-%5Ccfrac%7B1%7D%7B3%7D%28x%2B1%29%20%5C%5C%5C%5C%5C%5C%20y-3%3D-%5Ccfrac%7B1%7D%7B3%7Dx-%5Ccfrac%7B1%7D%7B3%7D%5Cimplies%20y%3D-%5Ccfrac%7B1%7D%7B3%7Dx-%5Ccfrac%7B1%7D%7B3%7D%2B3%5Cimplies%20y%3D-%5Ccfrac%7B1%7D%7B3%7Dx%2B%5Ccfrac%7B8%7D%7B3%7D)
Answer:
All of them are yes because you either subtract or add to put in the form of mx+b
Answer:
8 and 12
Step-by-step explanation:
Sides on one side of the angle bisector are proportional to those on the other side. In the attached figure, that means
AC/AB = CD/BD = 2/3
The perimeter is the sum of the side lengths, so is ...
25 = AB + BC + AC
25 = AB + 5 + (2/3)AB . . . . . . substituting AC = 2/3·AB. BC = 2+3 = 5.
20 = 5/3·AB
12 = AB
AC = 2/3·12 = 8
_____
<em>Alternate solution</em>
The sum of ratio units is 2+3 = 5, so each one must stand for 25/5 = 5 units of length.
That is, the total of lengths on one side of the angle bisector (AC+CD) is 2·5 = 10 units, and the total of lengths on the other side (AB+BD) is 3·5 = 15 units. Since 2 of the 10 units are in the segment being divided (CD), the other 8 must be in that side of the triangle (AC).
Likewise, 3 of the 15 units are in the segment being divided (BD), so the other 12 units are in that side of the triangle (AB).
The remaining sides of the triangle are AB=12 and AC=8.