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erastovalidia [21]
2 years ago
9

Given the function defined in the table below, find the average rate of change, in

Mathematics
1 answer:
yaroslaw [1]2 years ago
5 0

Answer:

Given the function defined in the table below, find the average rate of change, in simplest form, of the function over the interval 0 < x ..

Step-by-step explanation:

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Mr. Kohl has a breaker containing n milliliters of solution to distribute to the students in his chemistry class. If he gives ea
77julia77 [94]

Answer:

There were 26 students in his class and the teacher had 83 ml of the solution.

Step-by-step explanation:

Mr. Kohl has a "x" amount of solution, if he divides it by the number of students "n" he'll give each student 3 milliliters  and have a left over of 5 milliliters. If the amount of solution Mr. Kohl had was "x + 21" then he'd be able to give each student 4 milliliters of the solution. From these informations we have:

x = 3*n + 5

(x + 21)/n = 4

x + 21 = 4*n

x = 4*n - 21

Now that we have two equations and two variables we can solve the system of equations, as seen bellow:

3*n + 5 = 4*n - 21

3*n - 4*n = -21 - 5

-n = -26

n = 26

x = 4*26 - 21 = 83 ml

There were 26 students in his class and the teacher had 83 ml of the solution.

3 0
3 years ago
Choose the graphic representation of the quadratic function f(x)=x^2-8x+24
grigory [225]
The Answer is ----D----

8 0
3 years ago
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2 While solving the equation 4(x+2) = 28, Becca
mrs_skeptik [129]
Distributive, if you do 4 times x and 4 times 2 you get 4x+8
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3 years ago
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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
Compute using long division: 1,234÷68
Kruka [31]

Answer:

Quotient = 18

Remainder = 10

Step-by-step explanation:

1234/68

=> 68 x 1 = 68

=> 123 - 68 = 55

=> Take the 4 down

=> 554/68

=> 68 x 8 = 544

=> 554 - 544  = 10

So, the quotient = 18.

Remainder = 10

4 0
3 years ago
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