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Llana [10]
3 years ago
6

How do I do this ??

Mathematics
2 answers:
Bingel [31]3 years ago
6 0

Answer:

y would be at 3

Step-by-step explanation:

  1. How much space is in between x and z?: 8 - -2 = 8 + 2 = 10
  2. The midpoint of 10 is 5: -2 + 5 = 3

I hope this helps!

Nostrana [21]3 years ago
4 0

Answer:

If I'm not mistaking, they want you to find the number in the middle that's between X and Z.

That number would be 3.

Hope this helps a bit! :)

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Answer:

Step-by-step explanation:

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3 years ago
Can someone explain this step by step? I'm very confused on how to solve it. Thanks!
Rashid [163]

Answer:

x = -8

Step-by-step explanation:

I will show each step algebraically. If you need further explanation, feel free to ask questions :)

\frac{x}{2} +7=3\\                     Rewrite

\frac{x}{2} =-4                        Subtract 7

x =-8                        Multiply by reciprocal (2/1)

I hope this helps!

3 0
2 years ago
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What is the reciprocal of 8 5/6​
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3. Determine whether each relationship represents a function. Explain your reasoning.
topjm [15]
In order for something to be a function, it has to pass the vertical line test, atleast in the standard coordinate plane. This is because a single x value can’t correlate with two separate y values. So the first one is not and the second one is
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Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______
Ludmilka [50]

Answer:

(a)\ \sec^2(\theta) = 82

(b)\ \cot(\theta) = \frac{1}{9}

(c)\ \cot(\frac{\pi}{2} - \theta) = 9

(d)\ \csc^2(\theta) = \frac{82}{81}

Step-by-step explanation:

Given

\tan(\theta) = 9

Required

Solve (a) to (d)

Using tan formula, we have:

\tan(\theta) = \frac{Opposite}{Adjacent}

This gives:

\frac{Opposite}{Adjacent} = 9

Rewrite as:

\frac{Opposite}{Adjacent} = \frac{9}{1}

Using a unit ratio;

Opposite = 9; Adjacent = 1

Using Pythagoras theorem, we have:

Hypotenuse^2 = Opposite^2 + Adjacent^2

Hypotenuse^2 = 9^2 + 1^2

Hypotenuse^2 = 81 + 1

Hypotenuse^2 = 82

Take square roots of both sides

Hypotenuse =\sqrt{82}

So, we have:

Opposite = 9; Adjacent = 1

Hypotenuse =\sqrt{82}

Solving (a):

\sec^2(\theta)

This is calculated as:

\sec^2(\theta) = (\sec(\theta))^2

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

Where:

\cos(\theta) = \frac{Adjacent}{Hypotenuse}

\cos(\theta) = \frac{1}{\sqrt{82}}

So:

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2

\sec^2(\theta) = (\sqrt{82})^2

\sec^2(\theta) = 82

Solving (b):

\cot(\theta)

This is calculated as:

\cot(\theta) = \frac{1}{\tan(\theta)}

Where:

\tan(\theta) = 9 ---- given

So:

\cot(\theta) = \frac{1}{\tan(\theta)}

\cot(\theta) = \frac{1}{9}

Solving (c):

\cot(\frac{\pi}{2} - \theta)

In trigonometry:

\cot(\frac{\pi}{2} - \theta) = \tan(\theta)

Hence:

\cot(\frac{\pi}{2} - \theta) = 9

Solving (d):

\csc^2(\theta)

This is calculated as:

\csc^2(\theta) = (\csc(\theta))^2

\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2

Where:

\sin(\theta) = \frac{Opposite}{Hypotenuse}

\sin(\theta) = \frac{9}{\sqrt{82}}

So:

\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2

\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2

\csc^2(\theta) = \frac{82}{81}

4 0
3 years ago
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