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Reptile [31]
2 years ago
15

The students at Midtown Middle school sold flowers as a fundraiser in September and October. In October, they charged $1.50 for

each flower. The October price was a 20% increase of the September price. Part A: What was the price of the flowers in September? Part B: The seventh-grade class earned 40% of the selling price of each flower. In September, they sold 900 flowers. In October, they sold 700 flowers. Did they earn more money in September or October? How much more?
Mathematics
1 answer:
swat322 years ago
7 0
Let the price of flowers in September be x, then (100 + 20)/100 * x = $1.50
120/100 * x = $1.50
1.2x = $1.50
x = $1.50/1.2 = $1.25
The price of flowers in September is $1.25

Total money realised in September = $1.25 x 900 = $1,125
40% earned = 0.4 x $1,125 = $450

Total money realised in October = $1.50 x 700 = $1,050
40% earned = 0.4 x $1,050 = $420

The class earned more money in September and they earned $30 more than in October.
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The square roots of 49·i in ascending order are;

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Step-by-step explanation:

The square root of complex numbers 49·i is found as follows;

x + y·i = r·(cosθ + i·sinθ)

Where;

r = √(x² + y²)

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Therefore;

49·i = 0 + 49·i

Therefore, we have;

r = √(0² + 49²) = 49

θ = arctan(49/0) → 90°

Therefore, we have;

49·i = 49·(cos(90°) + i·sin(90°)

By De Moivre's formula, we have;

r \cdot (cos(\theta) + i \cdot sin(\theta) )^{\dfrac{1}{2}}  =  \pm \sqrt{r} \cdot \left (cos\left (\dfrac{\theta}{2} \right ) + i \cdot sin\left (\dfrac{\theta}{2} \right ) \right )

Therefore;

√(49·i) = √(49·(cos(90°) + i·sin(90°)) = ± √49·(cos(90°/2) + i·sin(90°/2))

∴ √(49·i) = ± √49·(cos(90°/2) + i·sin(90°/2)) = ± 7·(cos(45°) + i·sin(45°))

√(49·i) = ± 7·(cos(45°) + i·sin(45°))

The square roots of 49·i in ascending order are;

√(49·i) = - 7·(cos(45°) + i·sin(45°))  and  7·(cos(45°) + i·sin(45°))

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