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Elden [556K]
2 years ago
12

PLEASE HELP ASAP DUE IN 3 MINUTES

Mathematics
1 answer:
Inga [223]2 years ago
7 0

Answer:

<u>1728 in²</u>

Step-by-step explanation:

For the area of B, you need to find the length and width (height and base, whatever you want to call it).

For the height, you would need to do: 144 - 36 - 24 - 36 = <u>48 inches</u>

For the base, you would need to do: 60 - 24 = <u>36 inches</u>

Now you multiply the base and height to get the area: 48 x 36 = <u>1728 in²</u>

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Use completing the square to solve the equation x^2+16x=-44.
Sidana [21]

Step-by-step explanation:

our equation is x²+16x = -44

  • x²+16x= -44
  • x² is the first term so weill have in the middle 2*x* a number
  • x²+2*x*8 = -44
  • the third term is 8² wich is 64 so we will add it in both sides
  • x²+2*x*8+64 = -44+64
  • (x+8)² = 20

Now that we have completed the perfect square let's solve the equation

  • (x+8)² = 20
  • x+8 = \sqrt{20}or x+8= -\sqrt{20}
  • x = -8+\sqrt{20} or x =  -8- \sqrt{20}

so the first answer is the correct one

64; 8 +/- \sqrt{20}

7 0
3 years ago
Read 2 more answers
Kindly solve this question
Dmitry_Shevchenko [17]

Answer:

x=-1/7

Step-by-step explanation:

\frac{3x-1}{3} -\frac{2x}{x-1} =x

\frac{(3x-1)(x-1)}{3x-3} -\frac{2x(3)}{3x-3} =x

\frac{3x^2-4x+1}{3x-3} -\frac{6x}{3x-3} =x

\frac{3x^2-10x+1}{3x-3} =x

\frac{3x^2-10x+1}{3x-3} =\frac{x(3x-3)}{3x-3}

\frac{3x^2-10x+1}{3x-3} =\frac{3x^2-3x}{3x-3}

3x^2-10x+1=3x^2-3x

-10x+1=-3x

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3 0
2 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

6) w₁= 5 - 3i w₂= -5 + 3i

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

6 0
3 years ago
Can someone help me? Thanks! :)
attashe74 [19]

Answer:

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7 0
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Rule:

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