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kvv77 [185]
3 years ago
15

4x - 2 + y = 6 - 2x for x

Mathematics
1 answer:
Vlada [557]3 years ago
8 0

Answer:

4x

Step-by-step explanation:

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Whats 21/40 in equivalent decimals​
GREYUIT [131]
The equivalent decimals for 21/40 is 0.525
5 0
3 years ago
What is the approximate area of the circle shown below 9.4 cm
aivan3 [116]

Answer:

Step-by-step explanation:

A= πr²

A= 22/7 multiply by the radius.

Divide 9.4cm by 2.

Whatever you get is your answer. Now take the formula I have shown you above. Whatever you get is your answer.

6 0
3 years ago
The graph of the piecewise function f(x) is shown.<br> 1 f(x)
Tema [17]

Answer:

In a word processing document or on a separate piece of paper, use the guide to construct a two-column proof proving that lines l and m are parallel.

6 0
2 years ago
Suppose you have two urns with poker chips in them. Urn I contains two red chips and four white chips. Urn II contains three red
Neporo4naja [7]

Answer:

Multiple answers

Step-by-step explanation:

The original urns have:

  1. Urn 1 = 2 red + 4 white = 6 chips
  2. Urn 2 = 3 red + 1 white = 4 chips

We take one chip from the first urn, so we have:

The probability of take a red one is : \frac{1}{3} (2 red from 6 chips(2/6=1/2))

For a white one is: \frac{2}{3}(4 white from 6 chips(4/6=(2/3))

Then we put this chip into the second urn:

We have two possible cases:

  • First if the chip we got from the first urn was white. The urn 2 now has 3 red + 2 whites = 5 chips
  • Second if the chip we got from the first urn was red. The urn two now has 4 red + 1 white = 5 chips

If we select a chip from the urn two:

  • In the first case the probability of taking a white one is of:  \frac{2}{5} = 40%  ( 2 whites of 5 chips)
  • In the second case the probability of taking a white one is of:  \frac{1}{5} = 20%  ( 1 whites of 5 chips)

This problem is a dependent event because the final result depends of the first chip we got from the urn 1.

For the fist case we multiply :

\frac{4}{6} x \frac{2}{5} = \frac{4}{15} = 26.66%   ( \frac{4}{6} the probability of taking a white chip from the urn 1, \frac{2}{5}  the probability of taking a white chip from urn two)

For the second case we multiply:

\frac{1}{3} x \frac{1}{5} = \frac{1}{30} = .06%   ( \frac{1}{3} the probability of taking a red chip from the urn 1, \frac{1}{5}   the probability of taking a white chip from the urn two)

8 0
3 years ago
There are 3 islands A,B,C. Island B is east of island A, 8 miles away. Island C is northeast of A, 5 miles away and northwest of
Nostrana [21]

Answer:

The bearing needed to navigate from island B to island C is approximately 38.213º.

Step-by-step explanation:

The geometrical diagram representing the statement is introduced below as attachment, and from Trigonometry we determine that bearing needed to navigate from island B to C by the Cosine Law:

AC^{2} = AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot \cos \theta (1)

Where:

AC - The distance from A to C, measured in miles.

AB - The distance from A to B, measured in miles.

BC - The distance from B to C, measured in miles.

\theta - Bearing from island B to island C, measured in sexagesimal degrees.

Then, we clear the bearing angle within the equation:

AC^{2}-AB^{2}-BC^{2}=-2\cdot AB\cdot BC\cdot \cos \theta

\cos \theta = \frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC}

\theta = \cos^{-1}\left(\frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC} \right) (2)

If we know that BC = 7\,mi, AB = 8\,mi, AC = 5\,mi, then the bearing from island B to island C:

\theta = \cos^{-1}\left[\frac{(7\mi)^{2}+(8\,mi)^{2}-(5\,mi)^{2}}{2\cdot (8\,mi)\cdot (7\,mi)} \right]

\theta \approx 38.213^{\circ}

The bearing needed to navigate from island B to island C is approximately 38.213º.

8 0
3 years ago
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