(2√5 + 3(√7))^2
(2√5 + 3(√7))(2√5 + 3(√7))
4*5 + 6√35 + 6√35 + 9*7
20 + 12√35 + 63
20 + 63 + 12√35
83 + 12√35
solution:
Z1 = 5(cos25˚+isin25˚)
Z2 = 2(cos80˚+isin80˚)
Z1.Z2 = 5(cos25˚+isin25˚). 2(cos80˚+isin80˚)
Z1.Z2 = 10{(cos25˚cos80˚ + isin25˚cos80˚+i^2sin25˚sin80˚) }
Z1.Z2 =10{(cos25˚cos80˚- sin25˚sin80˚+ i(cos25˚sin80˚+sin25˚cos80˚))}
(i^2 = -1)
Cos(A+B) = cosAcosB – sinAsinB
Sin(A+B) = sinAcosB + cosAsinB
Z1.Z2 = 10(cos(25˚+80˚) +isin(25˚+80˚)
Z1.Z2 = 10(cos105˚+ isin105˚)
6b+2 I think I combined them and that’s what I got
Andrew can buy maximum 3 meals this weekend.
From given question,
Andrew must spend less than 53$ on meals during the weekend.
He has already spent 21$ on meals costing 8$ average.
Let x, the number of meals
So, we get an inequality,
8x + 21 < 53
We need to find the number of meals he can buy this weekend.
From above inequality,
⇒ 8x + 21 < 53
⇒ 8x < 53 - 21
⇒ 8x < 32
⇒ x < 4
This means, from 1 to 3 meals.
Therefore, Andrew can buy maximum 3 meals this weekend.
Learn more about an inequality here:
brainly.com/question/19003099
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