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3 years ago

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How does graphing -3/4 different from graphing 3/4

1 answer:

katovenus [111]3 years ago

3 0

Graph 1: y= -3/4, which means its (x,y) coordinate is (0, -3/4).

Graph 2: y= 3/4, which means its (x,y) coordinate is (0, 3/4).

Graph 1 would be the point -3/4 on the y-axis, while graph 2 would be the point 3/4 on the y-axis.

Therefore, graph 1 and graph 2 are symmetrical about the x-axis.

Hope this would help~

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What is x for the following:<br><br> |x-1/x-1|=1<br><br> and...<br><br> x-1/x-1

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Answer:

2, or any negative number

Step-by-step explanation:

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4 years ago

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Samantha sends her son, Barry, to a preschool center on certain days. The cost of preschool is $45 per day along with a fixed mo

Travka [436]

Answer: Option D

D. $880 = 45d + 70$ ; $18\ days$

Step-by-step explanation:

We know that the cost of preschool is $ 45 per day plus a monthly fee of $ 70.

We also know that a total of $ 880 was paid last month

To write an equation that represents this situation, let us call "d" the number of days that Barry attends school

So the cost was:

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3 years ago

Which inequality represents all values of X for which the product below is defined

balandron [24]

The domain for which the function is defined is given by:

D. $x \geq 4$ .

<h3>What is the domain of a function?</h3>

The domain of a function is the set that contains all possible input values for the function.

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More can be learned about the domain of a function at brainly.com/question/10891721

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2 years ago

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:

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

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

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

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3 years ago

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EleoNora [17]

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3 years ago