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2 years ago

Solve P=2 (1+w) for 1.

1 answer:

6 0

P = 2 ( 1 + w ) for 1

1 = 2 ( 1 + w )

Simplify

1 = 2w + 2

Subtract 2 from both sides

2w = -1

Divide by 2

2w = -1
/2 /2

w = -1/2

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A right triangle has a leg of 11 and a hypotenuse of 16. Find the missing leg (in simplest radical form)

wel

Answer: d

Step-by-step explanation:

6 0

2 years ago

Math question down below

SIZIF [17.4K]

To get the average, you must add all together and divide by the amount of numbers there are

(-4 - 6 - 3 - 1 + 0 + 2 - 1 - 3)/ 8

(-16)/8 = -2

-2 is your average

hope this helps

5 0

3 years ago

Read 2 more answers

Let A = {a, b, c}, B = {b, c, d}, and C = {b, c, e}. (a) Find A ∪ (B ∩ C), (A ∪ B) ∩ C, and (A ∪ B) ∩ (A ∪ C). (Enter your answe

wariber [46]

Answer:

(a)

$A\ u\ (B\ n\ C) = \{a,b,c\}$

$(A\ u\ B)\ n\ C = \{b,c\}$

$(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}$

$(A\ u\ B)\ n\ C = (A\ u\ B)\ n\ (A\ u\ C)$

(b)

$A\ n\ (B\ u\ C) = \{b,c\}$

$(A\ n\ B)\ u\ C = \{b,c,e\}$

$(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}$

$A\ n\ (B\ u\ C) = (A\ n\ B)\ u\ (A\ n\ C)$

(c)

$(A - B) - C = \{a\}$

$A - (B - C) = \{a,b,c\}$

They are not equal

Step-by-step explanation:

Given

$A= \{a,b,c\}$

$B =\{b,c,d\}$

$C = \{b,c,e\}$

Solving (a):

$A\ u\ (B\ n\ C)$

$(A\ u\ B)\ n\ C$

$(A\ u\ B)\ n\ (A\ u\ C)$

$A\ u\ (B\ n\ C)$

B n C means common elements between B and C;

So:

$B\ n\ C = \{b,c,d\}\ n\ \{b,c,e\}$

$B\ n\ C = \{b,c\}$

So:

$A\ u\ (B\ n\ C) = \{a,b,c\}\ u\ \{b,c\}$

u means union (without repetition)

So:

$A\ u\ (B\ n\ C) = \{a,b,c\}$

Using the illustrations of u and n, we have:

$(A\ u\ B)\ n\ C$

$(A\ u\ B)\ n\ C = (\{a,b,c\}\ u\ \{b,c,d\})\ n\ C$

Solve the bracket

$(A\ u\ B)\ n\ C = (\{a,b,c,d\})\ n\ C$

Substitute the value of set C

$(A\ u\ B)\ n\ C = \{a,b,c,d\}\ n\ \{b,c,e\}$

Apply intersection rule

$(A\ u\ B)\ n\ C = \{b,c\}$

$(A\ u\ B)\ n\ (A\ u\ C)$

In above:

$A\ u\ B = \{a,b,c,d\}$

Solving A u C, we have:

$A\ u\ C = \{a,b,c\}\ u\ \{b,c,e\}$

Apply union rule

$A\ u\ C = \{b,c\}$

So:

$(A\ u\ B)\ n\ (A\ u\ C) = \{a,b,c,d\}\ n\ \{b,c\}$

$(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}$

The equal sets

We have:

$A\ u\ (B\ n\ C) = \{a,b,c\}$

$(A\ u\ B)\ n\ C = \{b,c\}$

$(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}$

So, the equal sets are:

$(A\ u\ B)\ n\ C$ and $(A\ u\ B)\ n\ (A\ u\ C)$

They both equal to $\{b,c\}$

So:

$(A\ u\ B)\ n\ C = (A\ u\ B)\ n\ (A\ u\ C)$

Solving (b):

$A\ n\ (B\ u\ C)$

$(A\ n\ B)\ u\ C$

$(A\ n\ B)\ u\ (A\ n\ C)$

So, we have:

$A\ n\ (B\ u\ C) = \{a,b,c\}\ n\ (\{b,c,d\}\ u\ \{b,c,e\})$

Solve the bracket

$A\ n\ (B\ u\ C) = \{a,b,c\}\ n\ (\{b,c,d,e\})$

Apply intersection rule

$A\ n\ (B\ u\ C) = \{b,c\}$

$(A\ n\ B)\ u\ C = (\{a,b,c\}\ n\ \{b,c,d\})\ u\ \{b,c,e\}$

Solve the bracket

$(A\ n\ B)\ u\ C = \{b,c\}\ u\ \{b,c,e\}$

Apply union rule

$(A\ n\ B)\ u\ C = \{b,c,e\}$

$(A\ n\ B)\ u\ (A\ n\ C) = (\{a,b,c\}\ n\ \{b,c,d\})\ u\ (\{a,b,c\}\ n\ \{b,c,e\})$

Solve each bracket

$(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}\ u\ \{b,c\}$

Apply union rule

$(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}$

The equal set

We have:

$A\ n\ (B\ u\ C) = \{b,c\}$

$(A\ n\ B)\ u\ C = \{b,c,e\}$

$(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}$

So, the equal sets are:

$A\ n\ (B\ u\ C)$ and $(A\ n\ B)\ u\ (A\ n\ C)$

They both equal to $\{b,c\}$

So:

$A\ n\ (B\ u\ C) = (A\ n\ B)\ u\ (A\ n\ C)$

Solving (c):

$(A - B) - C$

$A - (B - C)$

This illustrates difference.

$A - B$ returns the elements in A and not B

Using that illustration, we have:

$(A - B) - C = (\{a,b,c\} - \{b,c,d\}) - \{b,c,e\}$

Solve the bracket

$(A - B) - C = \{a\} - \{b,c,e\}$

$(A - B) - C = \{a\}$

Similarly:

$A - (B - C) = \{a,b,c\} - (\{b,c,d\} - \{b,c,e\})$

$A - (B - C) = \{a,b,c\} - \{d\}$

$A - (B - C) = \{a,b,c\}$

They are not equal

4 0

3 years ago

Write an augmented matrix of each system. Then, solve each system using matrix notation. Describe the solution set in vector not

Naddik [55]

Answer:

The answer is below

Step-by-step explanation:

Given the system of equations:

x + y + 2z = 9

2x + 4y - 3z = 1

3x + 6y - 5z = 0

This system of equation can be solved using matrix. This done by first representing the equations as matrix and then solving:

The matrix form is:

$\left[\begin{array}{ccc}1&1&2\\2&4&-3\\3&6&-5\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}9\\1\\0\end{array}\right] \\\\\\ \left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}1&1&2\\2&4&-3\\3&6&-5\end{array}\right] ^{-1} \left[\begin{array}{c}9\\1\\0\end{array}\right] \\\\\\$

$\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}2&-17&11\\-1&11&-7\\0&3&-2\end{array}\right] \left[\begin{array}{c}9\\1\\0\end{array}\right] \\\\\\ \left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{c}1\\2\\3\end{array}\right]$