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pychu [463]
3 years ago
14

Solve: | −2n | + 10 = −50 (Find Solution)

Mathematics
1 answer:
CaHeK987 [17]3 years ago
4 0

Answer:

No Solution

Step-by-step explanation:

| −2n | + 10 = −50

Subtract 10 on both sides

|-2n|+10-10=-50-10

Simplify

|-2n=|=-60

No Solution

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7 cos x+1=6 secx solve for x
jonny [76]
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7cos(x) + 1 = 6/cos(x)
7cos^(x) + cos(x) = 6
7cos^(x) + cos(x) - 6 = 0
[7cos(x) - 6][cos(x) + 1] = 0
cos(x) = 6/7 , x = arccos(6/7) and
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6 0
2 years ago
HELP PLZ Which confidence level would produce the widest interval when estimating
Nikolay [14]

Answer:

83%

Step-by-step explanation:

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3 years ago
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Using the bijection rule to count binary strings with even parity.
AleksandrR [38]

Answer:

Lets denote c the concatenation of strings. For a binary string <em>a</em> in B9, we define the element f(a) in E10 this way:

  • f(a) = a c {1} if a has an odd number of 1's
  • f(a) = a c {0} if a has an even number of 1's

Step-by-step explanation:

To show that the function f defined above is a bijective function, we need to prove that f is well defined, injective and surjective.

f   is well defined:

To see this, we need to show that f sends elements fromo b9 to elements of E10. first note that f(a) has 1 more binary integer than a, thus, it has 10. if a has an even number of 1's, then f(a) also has an even number because a 0 was added. On the other hand, if a has an odd number of 1's, then f(a) has one more 1, as a consecuence it will have an even number of 1's. This shows that, independently of the case, f(a) is an element of E10. Thus, f is well defined.

f is injective (or one on one):

If a and b are 2 different binary strings, then f(a) and f(b) will also be different because the first 9 elements of f(a) form a and the first elements of f(b) form b, thus f(a) is different from f(b). This proves that f in injective.

f is surjective:

Let y be an element of E10, Let x be the first 9 elements of y, then f(x) = y:

  • If x has an even number of 1's, then the last digit of y has to be 0, and f(x) = x c {0} = y
  • If x has an odd number of 1's, then the last digit of y has to be a 1, otherwise it wont be an element of E10, and f(x) = x c {1} = y

This shows that f is well defined from B9 to E10, injective, and surjective, thus it is a bijection.

3 0
3 years ago
Given matrices A and B shown below, find (3)A+(8)B.
Tasya [4]

Answer:

\left[\begin{array}{ccc}32&-55\\3&39\end{array}\right]

Step-by-step explanation:

\left[\begin{array}{ccc}0&-15\\3&15\end{array}\right] +\left[\begin{array}{ccc}32&-40\\0&24\end{array}\right] =\left[\begin{array}{ccc}0+32&-15+(-40)\\3+0&15+24\end{array}\right] = \left[\begin{array}{ccc}32&-55\\3&39\end{array}\right]

5 0
2 years ago
*PLEASE HELP WILL GIVE 60 POINTS I'M DESPERATE*
Zina [86]

Answer:

(0, 3)

Step-by-step explanation:

1. Find the slope:

\frac{y_{2} -y_{1} }{x_{2}-x_{1}  } =\frac{0-(-3)}{-4-(-8)}\\ \\=\frac{3}{4}

2. Calculate the y-intercept using a given point:

(-8, -3)

-3 = \frac{3}{4} (-8) + b

-3 = -6 + b

-3+6 = -6+6 + b

3 = b

3. Write in slope intercept form:

y= \frac{3}{4} x+3

Therefore, the y-intercept is 3.

The graph for this table is shown below.

hope this helps :)

4 0
2 years ago
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