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Ksivusya [100]
3 years ago
12

I don’t understand the idea how do I do it??

Mathematics
1 answer:
vovangra [49]3 years ago
8 0
Independent variable = $20

dependent variable = $0.05 cents per min

y = total month cost
x = month minutes used
y = 20 + 0.05x
You might be interested in
I don't know what the answer is to divide 180 in a ratio of 4:5:9
Karolina [17]
Hello there,

First you need to add together the numbers.
4+5+9 = 18
Divide the total by this,
180 / 18 = 10
Multiply each number by 10.
40:50:90.

Hope This Helps You!
Good Luck Studying :)
4 0
3 years ago
Read 2 more answers
Find the range of the dataset<br> 8,11,5,22,9,18,5,20
lakkis [162]

Answer: 17

Step-by-step explanation: The range of a data set is the difference between the largest and smallest numbers

In this data set, the largest number is 22 and the smallest number is 5 so the range of the data set will simply be 22 - 5 or 17.

So the range of this data set is 17.

3 0
3 years ago
Read 2 more answers
The formula for the area of a square is s2, where s is the side length of the square. What is the area of a square with a side l
levacccp [35]

36

area of square = s²

given s = 6 , then

area = 6² = 36



8 0
3 years ago
Please help ASAP!!!!!!!!!
algol [13]

Answer:

A = 23.6 units^2

Step-by-step explanation:

Let the base be 8 (as shown).

Find the height of the triangle:   Find the supplement of the 100 degree angle; it is (180 - 100), or 80 degrees.  The side "opposite" this 80-degree angle is the height of the triangle:

   

                               height      

sin 80 degrees = ---------------

                                    6

and so the height of the triangle is h = 6 sin 80 degrees, or 5.91 units.

The area of the triangle is found using A = (1/2)(base)(height), which here amounts to:

A = (1/2)(8 units)(5.91 units), or

                                                       A = 23.64 units^2, or A = 23.6 units^2

3 0
3 years ago
I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).
Aleksandr-060686 [28]

Answer:

\frac{7\pi}{24} and \frac{31\pi}{24}

Step-by-step explanation:

\sqrt{3} \tan(x-\frac{\pi}{8})-1=0

Let's first isolate the trig function.

Add 1 one on both sides:

\sqrt{3} \tan(x-\frac{\pi}{8})=1

Divide both sides by \sqrt{3}:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

Now recall \tan(u)=\frac{\sin(u)}{\cos(u)}.

\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}

or

\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}

The first ratio I have can be found using \frac{\pi}{6} in the first rotation of the unit circle.

The second ratio I have can be found using \frac{7\pi}{6} you can see this is on the same line as the \frac{\pi}{6} so you could write \frac{7\pi}{6} as \frac{\pi}{6}+\pi.

So this means the following:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

is true when x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

where n is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

Add \frac{\pi}{8} on both sides:

x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi

Find common denominator between the first two terms on the right.

That is 24.

x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi

x=\frac{7\pi}{24}+n \pi (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval [0,2\pi).

So if \sqrt{3} \tan(x-\frac{\pi}{8})-1=0 and we let u=x-\frac{\pi}{8}, then solving for x gives us:

u+\frac{\pi}{8}=x ( I just added \frac{\pi}{8} on both sides.)

So recall 0\le x.

Then 0 \le u+\frac{\pi}{8}.

Subtract \frac{\pi}{8} on both sides:

-\frac{\pi}{8}\le u

Simplify:

-\frac{\pi}{8}\le u

-\frac{\pi}{8}\le u

So we want to find solutions to:

\tan(u)=\frac{1}{\sqrt{3}} with the condition:

-\frac{\pi}{8}\le u

That's just at \frac{\pi}{6} and \frac{7\pi}{6}

So now adding \frac{\pi}{8} to both gives us the solutions to:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}} in the interval:

0\le x.

The solutions we are looking for are:

\frac{\pi}{6}+\frac{\pi}{8} and \frac{7\pi}{6}+\frac{\pi}{8}

Let's simplifying:

(\frac{1}{6}+\frac{1}{8})\pi and (\frac{7}{6}+\frac{1}{8})\pi

\frac{7}{24}\pi and \frac{31}{24}\pi

\frac{7\pi}{24} and \frac{31\pi}{24}

5 0
3 years ago
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