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ivann1987 [24]
2 years ago
8

If r=1 and 0=5pie/3, what is the approximate arc length?​

Mathematics
1 answer:
Masja [62]2 years ago
4 0

Answer:

5.236 units

Step-by-step explanation:

The relation between r, θ and arc length ( l ) is,

l = rθ

l = 1 * 5π / 3

( The value of π = 3.14 approximately )

5π = 5 x 3.14 = 15.7

l = 1 * 5π / 3

= 5π / 3

= 15.7/3

l = 5.236 units

Note : -

Arc length is denoted by " l ".

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Q1- Write 46 821 correct to the nearest 100
noname [10]

Answer:

46800

Step-by-step explanation:

46821 nearest 100s

look for the tens it is less than 5 so its 800 and the numbers written before the 100s would remain the same

8 0
3 years ago
which of the functions have a range of real numbers greater than or equal to 1 or less than or equal to-1​
lilavasa [31]

The question is incomplete. Here is the complete question:

Which of the functions have a range of all real numbers greater than or equal to 1 or less than or equal to -1? check all that apply.

A. y=\sec x

B. y= \tan x

C. y= \cot x

D. y= \csc x

Answer:

A. y=\sec x

D. y=\csc x

Step-by-step explanation:

Given:

The range is greater than or equal to 1 or less than or equal to -1.

The given choices are:

Choice A: y=\sec x

We know that, the \sec x=\frac{1}{\cos x}

The range of \cos x is from -1 to 1 given as [-1, 1]. So,

|\cos x|\leq 1\\\textrm{Taking reciprocal, the inequality sign changes}\\\frac{1}{|\cos x|}\geq 1\\|\sec x|\geq 1

Therefore, on removing the absolute sign, we rewrite the secant function as:

\sec x\leq -1\ or\ \sec x\geq 1\\

Therefore, the range of y=\sec x is all real numbers greater than or equal to 1 or less than or equal to-1​.

Choice B: y= \tan x

We know that, the range of tangent function is all real numbers. So, choice B is incorrect.

Choice C: y= \cot x

We know that, the range of cotangent function is all real numbers. So, choice C is incorrect.

Choice D: y=\csc x

We know that, the \csc x=\frac{1}{\sin x}

The range of \sin x is from -1 to 1 given as [-1, 1]. So,

|\sin x|\leq 1\\\textrm{Taking reciprocal, the inequality sign changes}\\\frac{1}{|\sin x|}\geq 1\\|\csc x|\geq 1

Therefore, on removing the absolute sign, we rewrite the cosecant function as:

\csc x\leq -1\ or\ \csc x\geq 1\\

Therefore, the range of y=\csc x is all real numbers greater than or equal to 1 or less than or equal to-1​.

6 0
3 years ago
The fractions and decimals in order from least to greatest 7/48 0.5 0.75 9/16
katen-ka-za [31]

Answer:

7/48,0.5,9/16,0.75

Step-by-step explanation:

Hope this helps

6 0
3 years ago
Explain the difference between -<img src="https://tex.z-dn.net/?f=%5Csqrt%7B1%7D" id="TexFormula1" title="\sqrt{1}" alt="\sqrt{1
OLEGan [10]

Step-by-step explanation:

= -√1

= -1

= √1

= 1

Difference -√1 and √1 are the sign of the number, where -√1 is negative and √1 is positive

3 0
3 years ago
Help I’m stuck on this question I’ve been stuck on it for a while I can’t seem to figure it out please help for ten points
Naddika [18.5K]

Answer:

\frac{1}{25}

Step-by-step explanation:

I think what it is trying to say is that it wants the solution to multiplying all of those. The (...) simply means that it wants you to continue that pattern in what you are supposed to be multiplying, but stop at \frac{24}{25}.

That would mean you are technically supposed to be multiplying:

\frac{1}{2}  \frac{2}{3} \frac{3}{4} \frac{4}{5} \frac{5}{6} \frac{6}{7} \frac{7}{8} \frac{8}{9} \frac{9}{10} \frac{10}{11} \frac{11}{12} \frac{12}{13} \frac{13}{14} \frac{14}{15} \frac{15}{16} \frac{16}{17} \frac{17}{18} \frac{18}{19} \frac{19}{20} \frac{20}{21} \frac{21}{22} \frac{22}{23} \frac{23}{24} \frac{24}{25}

That is a lot and unfortunately, none of the individual fractions can be simplified within that. The final answer would be able to be simplified, though.

Multiplying the first four shown: \frac{1}{2}\frac{2}{3} \frac{3}{4} \frac{4}{5} you end up with \frac{24}{120}. Both the numerator (top) and the denominator (bottom) are divisible by 24. Dividing top and bottom would simplify this to \frac{1}{5}.

Now, let's take the next four.

\frac{5}{6}\frac{6}{7} \frac{7}{8} \frac{8}{9} allows you to end up with \frac{1680}{3024}. Both the numerator (top) and the denominator (bottom) are divisible by 336. You are left with \frac{5}{9}.

Now, let's take the next four.

\frac{9}{10}\frac{10}{11} \frac{11}{12} \frac{12}{13}. Multiplying these gives you \frac{11880}{17160}. Both the numerator (top) and the denominator (bottom) are divisible by 1320. You are left with \frac{9}{13}.

Now, let's take the next four.

\frac{13}{14}\frac{14}{15} \frac{15}{16} \frac{16}{17}. Multiplying these gives you \frac{43680}{57120}. Both the numerator (top) and the denominator (bottom) are divisible by 3360. You are left with \frac{13}{17}.

*<em> Although I would continue to say let's take the next four, there appears to be a pattern in the simplification. The numerators we have gotten have all been four less than their denominators, and each numerator has been four more than the last. I cannot be certain, but we only have two sets of four left. If this pattern continues, the simplifications of each should be \frac{17}{21} and \frac{21}{25}. I will continue on for argument's sake, anyways.</em>

The next four are \frac{17}{18}\frac{18}{19} \frac{19}{20} \frac{20}{21}. Multiplying these, you are left with \frac{116280}{143640}. Both the numerator (top) and the denominator (bottom) are divisible by 6840. We are left with\frac{17}{21}. This is the exact guess I had made when following the pattern, and so the next one is most likely going to be the other guess as well.

Our final answer will be \frac{1}{5}\frac{5}{9} \frac{9}{13} \frac{13}{17}\frac{17}{21} \frac{21}{25} all multiplied together. We end up with \frac{208845}{5221125}. Both the numerator (top) and denominator (bottom) are divisible by 208845.

Simplified, your final answer is:  \frac{1}{25}.

* <u>NOTE:</u> that another way to solve this would just be to multiply all numbers from 1-24 together and then 2-25, but you would end up with a very large number that would be just as time consuming to simplify. To get the GCF fast, I used a GCF calculator.

6 0
3 years ago
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