18,-2,-14 just plug in for x
Given that
(2+i)²/(3+i)
On multiplying both numerator and the denominator with (3-i) then
⇛ [(2+i)²/(3+i)]×[(3-i)/(3-i)]
⇛ [(2+i)²(3-i)]/[(3+i)(3-i)]
⇛ [(2+i)²(3-i)]/[(3²-i²)
⇛ [(2+i)²(3-i)]/(9-i²)
⇛ [(2+i)²(3-i)]/[9-(-1)]
Since ,i² = -1
⇛ [(2+i)²(3-i)]/(9+1)
⇛ [(2+i)²(3-i)]/10
⇛ [{2²+i²+2(2)(i)}(3-i)]/10
⇛ (4+i²+4i)(3-i)/10
⇛ (4-1+4i)(3-i)/10
⇛ (3+4i)(3-i)/10
⇛ (9-3i+12i-4i²)/10
⇛ (9+9i-4(-1))/10
Since, i² = -1
⇛(9+9i+4)/10
⇛(13+9i)/10
⇛ (13/10)+ i (9/10)
We know that
The conjugate of a+ib is a-ib
So,
The conjugate of (13/10)+ i (9/10) is
(13/10)-i(9/10) ⇛ (13/10)+i (-9/10)
<u>Answer:-</u>The conjugate of (13/10)+ i (9/10) is (13/10)+i (-9/10)
<em>Additional</em><em> comment</em><em>:</em>
- The conjugate of a+ib is a-ib and
- i = -1
- (a+b)² = a²+2ab+b² • (a+b)(a-b)=a²-b² •(a-b)²=a²-2ab+b².
C hope that helps have a good one and stay 28 - 6973
add the 2 angles: 89+80= 169
subtract 180-169= 11
11° is the measurement of the final angle
do sin(80)=12/x
x(sin(80))=12
plug in on calculator:
x=12/(sin(80))
The expression factorized completely is (h +2k)[(h+2k) + (2k-h)]
From the question,
We are to factorize the expression (h+2k)²+4k²-h² completely
The expression can be factorized as shown below
(h+2k)²+4k²-h² becomes
(h+2k)² + 2²k²-h²
(h+2k)² + (2k)²-h²
Using difference of two squares
The expression (2k)²-h² = (2k+h)(2k-h)
Then,
(h+2k)² + (2k)²-h² becomes
(h+2k)² + (2k +h)(2k-h)
This can be written as
(h+2k)² + (h +2k)(2k-h)
Now,
Factorizing, we get
(h +2k)[(h+2k) + (2k-h)]
Hence, the expression factorized completely is (h +2k)[(h+2k) + (2k-h)]
Learn more here:brainly.com/question/12486387
