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Reptile [31]
3 years ago
14

Steven, a tailor, got an order to make a blazer. The customer specifically asked him to save 5⁄6 of a foot of the given cloth to

make a pocket square. However, Steven accidentally saved 5⁄12 of a foot. What is the difference between the requested cloth and the saved cloth? A. 0.4265′ B. 0.1466′ C. 0.4166′ D. 0.4066′
Mathematics
1 answer:
xz_007 [3.2K]3 years ago
6 0
To answer this question, we just need to subtract the saved cloth from the requested cloth:

\frac{5}{6} - \frac{5}{12}

Since the fractions don't have the same denominator, we have to find the least common denominator.  Luckily, since 6 goes into 12 twice, we can convert \frac{5}{6} into \frac{10}{12}:

\frac{5}{6}( \frac{2}{2}) - \frac{5}{12} = \frac{10}{12} - \frac{5}{12}

Now that the fractions have a common denominator, we can subtract them:

\frac{10}{12} - \frac{5}{12} = \frac{5}{12}

The difference between the requested cloth and the saved cloth is 5/12 of a foot, or 0.4166'.
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kvasek [131]

Answer:

139, I am sorry if I'm wrong

Step-by-step explanation:

a1=11; d=19-11=8

a17=11+(17-1)*8=11+128=139

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Tracking a change over time.
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3 years ago
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Math:
Dahasolnce [82]

Answer:

a) x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i, \forall i \in \mathbb{N}_{O}, x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i, \forall i \in \mathbb{N}_{O}, b) x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i, \forall i \in \mathbb{N}_{O}, x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i, \forall i \in \mathbb{N}_{O}

Step-by-step explanation:

a) The equation must be rearranged into a form with one fundamental trigonometric function first:

\sqrt{3}\cdot \csc x - 2 = 0

\sqrt{3} \cdot \left(\frac{1}{\sin x} \right) - 2 = 0

\sqrt{3} - 2\cdot \sin x = 0

\sin x = \frac{\sqrt{3}}{2}

x = \sin^{-1} \frac{\sqrt{3}}{2}

Value of x is contained in the following sets of solutions:

x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i, \forall i \in \mathbb{N}_{O}

x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i, \forall i \in \mathbb{N}_{O}

b) The equation must be simplified first:

\cos x + 1 = - \cos x

2\cdot \cos x = -1

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

x = \cos^{-1} \left(-\frac{1}{2} \right)

Value of x is contained in the following sets of solutions:

x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i, \forall i \in \mathbb{N}_{O}

x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i, \forall i \in \mathbb{N}_{O}

7 0
3 years ago
Could someone confirm if I was correct or not?
Andrei [34K]
Correct! Excellent Job
7 0
2 years ago
Read 2 more answers
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