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3 years ago

14

Use the net to find the surface area of the regular pyramid.

1 answer:

pychu [463]3 years ago

7 0

Answer:

the answer would be 78.6

Step-by-step explanation:

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The table below shows values for functions f(x) = g(x) <br> What are the solutions to f(x) = g(x)?

MakcuM [25]

The solutions to f(x) = g(x) are where the x-values for which the output f(x) is equal to the output of g(x).

What I mean by this is for instance, you input 7 into f(x) and g(x) and you get the same answer, then 7 is a solution.

Here, we are looking in columns two and three to see which rows are equal. It looks like when you input 0 into both f(x) and g(x), you get 2, and when you input 1 into both f(x) and g(x), you get 3. Therefore, (0,2) and (1,3) are your solutions.

8 0

3 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:

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

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

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

4 0

3 years ago

Val needs to find the area enclosed by the figure. The figure is made by attaching semicircles to each side of a 58-by-58 squar

Savatey [412]

Answer:

We need to find the area of the semicircles + the area of the square.

The area of a square is equal to the square of the lenght of one side.

As = L^2 = 58m^2 = 3,364 m^2

Now, each of the semicircles has a diameter of 58m, and we have that the area of a circle is equal to:

Ac = pi*(d/2)^2 = 3.14*(58m/2)^2 = 3.14(27m)^2 = 2,289.06m^2

And the area of a semicircle is half of that, so the area of each semicircle is:

a = (2,289.06m^2)/2 = 1,144.53m^2

And we have 4 of those, so the total area of the semicircles is:

4*a = 4* 1,144.53m^2 = 4578.12m^2

Now, we need to add the area of the square 3,364 m^2 + 4578.12m^2 = 7942.12m^2

This is nothing like the provided anwer of Val, so the numbers of val may be wrong.

4 0

3 years ago

50 points! Look at the picture attached. I will mark brainliest

Radda [10]

I’m pretty sure the answer would be 116

8 0

3 years ago

Read 2 more answers

1 and 2 is all I need and you guys would help me from not getting beat

Sonja [21]

Answer:

1. -10

2. -11

Step-by-step explanation:

6 0

3 years ago