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Gwar [14]
3 years ago
9

Find two positive fractions with a quotient of 5/6

Mathematics
1 answer:
hichkok12 [17]3 years ago
3 0
Alright, so if we divide fractions we get
(x/y)/(a/b)= (x/y)*(b/a). If the x is 5 and the a is 6, as well as y and b being the same therefore cancelling out, we can write it easily. For example, we can write (5/100)/(6/100) or (5/60)/(6/60)
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Type SSS, SAS, ASA, SAA, or HL to<br> describe these triangles.
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SAS

Step-by-step explanation:

There is one common side (S)

both the triangles have 90° common (A)

Opposite sides are equal which is given (S)

They both are right angled triangles

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What is the simplest form of this expression?
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The answer is B.
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alice leaves her house and walks to school she walks 45 meters south and 336 meters east. how far is Alice from her house?
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The quotient of a number and 6decreased by 10 is equal to 2??
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Consider w=sqrrt2/2(cos(225°) + isin(225°)) and z = 1(cos(60°) + isin(60°)). What is w+ z expressed in rectangular form?
SVETLANKA909090 [29]

Answer:

Option (3)

Step-by-step explanation:

w = \frac{\sqrt{2}}{2}[\text{cos}(225) + i\text{sin}(225)]

Since, cos(225) = cos(180 + 45)

                          = -cos(45) [Since, cos(180 + θ) = -cosθ]

                          = -\frac{\sqrt{2}}{2}

sin(225) = sin(180 + 45)

             = -sin(45)

             = -\frac{\sqrt{2}}{2}

Therefore, w = \frac{\sqrt{2}}{2}[-\frac{\sqrt{2}}{2}+i(-\frac{\sqrt{2}}{2})]

                      = -\frac{2}{4}(1+i)

                      = -\frac{1}{2}(1+i)

z = 1[cos(60) + i(sin(60)]

  = [\frac{1}{2}+i(\frac{\sqrt{3}}{2})

  = \frac{1}{2}(1+i\sqrt{3})

Now (w + z) = -\frac{1}{2}(1+i)+\frac{1}{2}(1+i\sqrt{3})

                   = -\frac{1}{2}-\frac{i}{2}+\frac{1}{2}+i\frac{\sqrt{3}}{2}

                   = \frac{(i\sqrt{3}-i)}{2}

                   = \frac{(\sqrt{3}-1)i}{2}

Therefore, Option (3) will be the correct option.

3 0
3 years ago
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