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jekas [21]
3 years ago
15

Jake is buying a sweater with the price of 30 dollars the store is having a sell that will reduce the price by 30%. Sales tax of

5% will be added to the final purchase price what is jake total cost.
Mathematics
1 answer:
shusha [124]3 years ago
7 0

Answer:Jake's total cost is $22.05

Step-by-step explanation:

The initial price of the sweater is $30. The store is having a sell that will reduce the price by 30%. This means that there is a reduction in the original amount of the sweater by 30% which is expressed as

30/100 × 30 = $9

The new price of the sweater would be 30 - 9 = $21

Since a sales tax of 5% will be added to the final purchase price, the value of the tax would be

5/100 × 21 = 0.05×21 = $1.05

Jake's total cost would be

21 + 1.05 = $22.05

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Answer:

<h2><em><u>The answer is 483</u></em></h2>

Step-by-step explanation:

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3 years ago
Find the two square roots of each complex number by creating and solving polynomial equations.
son4ous [18]

Answer:

1) w₁=4 - i w₂= -4 + i

2) w₁= 3 - i w₂= -3  + i

3) w₁= 1 + 2i w₂= - 1 - 2i

4) w₁= 2- 3i w₂= -2 + 3i

5) w₁= 5 - 2i w₂= -5 + 2i

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Step-by-step explanation:

The root of a complex number is given by:

\sqrt[n]{z}=\sqrt[n]{r}(Cos(\frac{\theta+2k\pi}{n}) + i Sin(\frac{\theta+2k\pi}{n}))

where:

r: is the module of the complex number

θ: is the angle of the complex number to the positive axis x

n: index of the root

1) z = 15 − 8i  ⇒ r=17 θ= -0.4899 rad

w₁=\sqrt{17}(Cos(\frac{-0.4899}{2}) + i Sin(\frac{-0.4899}{2}))=4-i

w₂=\sqrt{17}(Cos(\frac{-0.4899+2\pi}{2}) + i Sin(\frac{-0.4899+2\pi}{2}))=-1+i

2) z = 8 − 6i  ⇒ r=10 θ= -0.6435 rad

w₁=\sqrt{10}(Cos(\frac{ -0.6435}{2}) + i Sin(\frac{ -0.6435}{2}))= 3 - i

w₂=\sqrt{10}(Cos(\frac{ -0.6435+2\pi}{2}) + i Sin(\frac{ -0.6435+2\pi}{2}))= -3  + i

3) z = −3 + 4i  ⇒ r=5 θ= -0.9316 rad

w₁=\sqrt{5}(Cos(\frac{-0.9316}{2}) + i Sin(\frac{-0.9316}{2}))= 1 + 2i

w₂=\sqrt{5}(Cos(\frac{-0.9316+2\pi}{2}) + i Sin(\frac{-0.9316+2\pi}{2}))= -1 - 2i

4) z = −5 − 12i  ⇒ r=13 θ= 0.4426 rad

w₁=\sqrt{13}(Cos(\frac{0.4426}{2}) + i Sin(\frac{0.4426}{2}))= 2- 3i

w₂=\sqrt{13}(Cos(\frac{0.4426+2\pi}{2}) + i Sin(\frac{0.4426+2\pi}{2}))= -2 + 3i

5) z = 21 − 20i  ⇒ r=29 θ= -0.8098 rad

w₁=\sqrt{29}(Cos(\frac{-0.8098}{2}) + i Sin(\frac{-0.8098}{2}))= 5 - 2i

w₂=\sqrt{29}(Cos(\frac{-0.8098+2\pi}{2}) + i Sin(\frac{-0.8098+2\pi}{2}))= -5 + 2i

6) z = 16 − 30i ⇒ r=34 θ= -1.0808 rad

w₁=\sqrt{34}(Cos(\frac{-1.0808}{2}) + i Sin(\frac{-1.0808}{2}))= 5 - 3i

w₂=\sqrt{34}(Cos(\frac{-1.0808+2\pi}{2}) + i Sin(\frac{-1.0808+2\pi}{2}))= -5 + 3i

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