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.
Answer:
Step-by-step explanation:
From the question attached,
Given:
RT and PQ intersect at a point S.
RS = PS and ST = SQ
To prove :
RT = PQ
Statements Reasons
1). RS = PS and ST = SQ 1). Given
2). RS + ST = PS + SQ 2). Addition property
3). RS + ST = RT ; PS + SQ = PQ 3). Pair of line segments
4). RT = PQ 4). Substitution property
Answer:
No solution.
Step-by-step explanation:
y + 3x - 4y = 0 + 0.75x
- 3y + 3x = 0 + 0.75x
- 3y + 3x - 0.75x = 0 + 0.75x - 0.75x
- 3y + 2.25x = 0
42/48 is equivalent to 7/8
