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Lady_Fox [76]
3 years ago
15

I need to do this now

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
8 0

Answer:

4.55294117. If you want it rounded look below.

Step-by-step explanation:

-Rounded-

Hundredths:4.55

Tenth: 4.6

Thousandth: 4.553

Whole Number: 5

Hope this helps. :D

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Please answer this correctly
valkas [14]

Answer:

all of them are valid solutions

Step-by-step explanation:

So we plug in values of t below and get:

18 is < 107

54 is < 107

36 is < 107

27 is <107

7 0
3 years ago
Please HELP ITS A GRADE AND TIMED I WILL MARK BRAINLIEST
qaws [65]

Answer:

Domain: All Real Numbers

Range: y<0

Step-by-step explanation:

Looking at the graph, you can tell that the variable y has a lower point.

( I'm not sure how accurate this is I'm doing the same test right now. ;-;)  

8 0
3 years ago
How many lines of symmetry does a regular polygon with 32 sides have
Leya [2.2K]
32. 16 through opposite vertices and 16 through the centres of opposite sides 
3 0
2 years ago
Help me asap please!!!
Artemon [7]
Well, just by looking at the beginning of the problem, Jamelia had made the common mistake of thinking that 
\sqrt{72} (<span>8.4852...)
</span>is equal to 
2* \sqrt{36} (12)

If you want to estimate a square root like 72, simply find squares that would fit around the number you are looking to find, in our case, 72.

So 9*9 is 81, which is too high and 8*8 is 64, which is too low. So you know that somewhere between those numbers is what your root of 72 is!
5 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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