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dangina [55]
3 years ago
5

How do you solve 3 q+2(q+1)

Mathematics
2 answers:
Liula [17]3 years ago
6 0
3q + 2 (q + 1)               :Distribute

3q + 2q + 1                  :Combine like terms

<em><u>5q + 1</u></em>
vova2212 [387]3 years ago
5 0
3q+2(q+1)=\ \ \ \ | omiting\ the\ bracket\\\\&#10;3q+2q+2=5q+2\\\\&#10;Solution\ is\ 5q+2.
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Answer:

140 liters will remain after 10 mins

Step-by-step explanation:

The easy way is just multiply the amount lost per min by the number of mins.

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Show the zero property of multiplication using the number 25
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Convert,the complex number into polar form: 4+4i
kow [346]
Z = a + bi
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r² = a² + b²
r² = (4)² + (4)²
r² = 16 + 16
r² = 32
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 r = 4(1.414)
 r = 5.656

cos\theta = \frac{a}{r}
cos\theta = \frac{4}{4\sqrt{2}}
cos\theta = \frac{4}{4\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}
cos\theta = \frac{4\sqrt{2}}{4\sqrt{4}}
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cos\theta = \frac{4\sqrt{2}}{8}
cos\theta = \frac{\sqrt{2}}{2}
2(cos\theta) = 2(\frac{\sqrt{2}}{2})
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sin\theta = \frac{b}{r}
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sin\theta = \frac{4}{4\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}
sin\theta = \frac{4\sqrt{2}}{4\sqrt{4}}
sin\theta = \frac{4\sqrt{2}}{4(2)}
sin\theta = \frac{4\sqrt{2}}{8}
sin\theta = \frac{\sqrt{2}}{2}
2(sin\theta) = 2(\frac{\sqrt{2}}{2})
2sin\theta = \sqrt{2}
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z = a + bi
z = rcosθ + (rsinθ)i
z = r(cosθ + i sinθ)

z = 4 + 4i
z = 5.656cosθ + (5.656sinθ)i
z = 5.656(cosθ + i sinθ)
z = 5.656(cos45 + i sin45)

\theta = tan^{-1}\frac{b}{a}
\theta = tan^{-1}\frac{4}{4}
\theta = tan^{-1}(1)
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The polar form of 4 + 4i is approximately equal to 5.656(cos45 + i sin45).
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3 years ago
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Answer:

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