![\boxed{\underline{\bf \: ANSWER}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cunderline%7B%5Cbf%20%5C%3A%20ANSWER%7D%7D)
![\sf \: ( 5 + 6 i ) ( 5 - 6 i )](https://tex.z-dn.net/?f=%20%5Csf%20%5C%3A%20%28%205%20%2B%206%20i%20%29%20%28%205%20-%206%20i%20%29)
Multiply complex numbers 5+6i and 5-6i in the same way as you multiply binomials.
![\dashrightarrow \sf 5\times 5+5\times \left(-6i\right)+6i\times 5+6\left(-6\right)i^{2}](https://tex.z-dn.net/?f=%20%5Cdashrightarrow%20%5Csf%205%5Ctimes%205%2B5%5Ctimes%20%5Cleft%28-6i%5Cright%29%2B6i%5Ctimes%205%2B6%5Cleft%28-6%5Cright%29i%5E%7B2%7D%20)
By definition, i² is -1.
![\dashrightarrow \sf 5\times 5+5\times \left(-6i\right)+6i\times 5+6\left(-6\right)\left(-1\right)](https://tex.z-dn.net/?f=%20%5Cdashrightarrow%20%5Csf%205%5Ctimes%205%2B5%5Ctimes%20%5Cleft%28-6i%5Cright%29%2B6i%5Ctimes%205%2B6%5Cleft%28-6%5Cright%29%5Cleft%28-1%5Cright%29%20)
Do the multiplications.
![\dashrightarrow \sf \: 25-30i+30i+36](https://tex.z-dn.net/?f=%20%5Cdashrightarrow%20%5Csf%20%5C%3A%2025-30i%2B30i%2B36%20)
Now cancel -30i & +30i
![\dashrightarrow \sf \: 25 \: \bcancel{- 30i} \: + \: \bcancel{30i}+36](https://tex.z-dn.net/?f=%5Cdashrightarrow%20%5Csf%20%5C%3A%2025%20%5C%3A%20%20%5Cbcancel%7B-%2030i%7D%20%20%5C%3A%20%20%2B%20%20%5C%3A%20%5Cbcancel%7B30i%7D%2B36%20)
Now the equation becomes
![\dashrightarrow \sf \: 25 \: + 36 \\ \dashrightarrow \boxed{\bf \: 61}](https://tex.z-dn.net/?f=%5Cdashrightarrow%20%5Csf%20%5C%3A%2025%20%5C%3A%20%20%2B%2036%20%5C%5C%20%5Cdashrightarrow%20%20%5Cboxed%7B%5Cbf%20%5C%3A%2061%7D)
________
Hope it helps.
ŕάίήвόώşάĻţ2²2²
42/98=3/7
Have a great day :)
9514 1404 393
Answer:
Step-by-step explanation:
In the left problem, you use the fact that <em>the sum of the segment lengths is equal to the overall length</em>.
AC +CB = AB
(3x -4) +(x -2) = 62
4x -6 = 62 . . . . . collect terms
4x = 68 . . . . . . . add 6
x = 17 . . . . . . . . . . divide by 4
__
In the right problem, you use the fact that <em>the sum of the angles is equal to the overall angle</em>. Here, that overall angle is a linear angle, so measures 180°.
∠DFG +∠GFE = ∠DFE
(5y +3) +(2y -5) = 180
7y = 182 . . . . . . . . . . . . . . collect terms, add 2
y = 26 . . . . . . . . . . . . . . . .divide by 7
Answer:
20 mph
Step-by-step explanation:
Let speed of Train a = Va mph
Let speed of Train b = Vb mph
Va = Vb + 30
Let time taken for Train a = Ta
Let time taken for Train b = Tb
time taken = distance travelled/speed
Ta = 350/Va = 350/(Vb+30)
Tb = 140/Vb
But they both travel in the same amount of time.
So, Ta = Tb
![\frac{350}{v_{b} + 30}=\frac{140}{v_{b}}](https://tex.z-dn.net/?f=%5Cfrac%7B350%7D%7Bv_%7Bb%7D%20%2B%2030%7D%3D%5Cfrac%7B140%7D%7Bv_%7Bb%7D%7D)
Cross multiply
![350v_{b} = 140(v_{b} + 30)\\\\350v_{b} = 140v_{b} + 4200\\\\350v_{b} -140v_{b} = 4200\\\\210v_{b} = 4200](https://tex.z-dn.net/?f=350v_%7Bb%7D%20%3D%20140%28v_%7Bb%7D%20%2B%2030%29%5C%5C%5C%5C350v_%7Bb%7D%20%3D%20140v_%7Bb%7D%20%2B%204200%5C%5C%5C%5C350v_%7Bb%7D%20-140v_%7Bb%7D%20%3D%204200%5C%5C%5C%5C210v_%7Bb%7D%20%3D%204200)
Divide both sides by 210
![\frac{210v_{b} }{210} \frac{4200}{210} \\\\v_{b} = 20 mph](https://tex.z-dn.net/?f=%5Cfrac%7B210v_%7Bb%7D%20%7D%7B210%7D%20%5Cfrac%7B4200%7D%7B210%7D%20%5C%5C%5C%5Cv_%7Bb%7D%20%3D%2020%20mph)
The most simplified version of it is 49/50