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Alenkasestr [34]
3 years ago
11

Two similar circles are shown. The circumference of the

Mathematics
1 answer:
yaroslaw [1]3 years ago
3 0

Answer:

(B). \frac{2\pi }{3} x

Step-by-step explanation:

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jeka94
There are at least two
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3 years ago
Order each set of numbers from least to greatest 0.99 0.89 7/8
kifflom [539]
7/8 = 0.875
1. 7/8
2. 0.89
3. 0.99 
3 0
3 years ago
Suzie made a mistake in the following problem. The mistake was made in Line ___. Only input the number of the first incorrect li
zalisa [80]

Answer:

Mistake in Line 2

Step-by-step explanation:

Line 1  

7(6) ÷ 5 + 42  

Line 2  

7(6) ÷ 5 + 16  

Line 3  

42 ÷ 5 + 16  

Line 4  

42 ÷ 21  

Line 5  

2

To simplify any expression we use order of operation

Line 1 is 7(6) ÷ 5 + 42  

Order of operation is PEMDAS

First we start with parenthesis

so we multiply 7(6) in line 2

7*6- 42

So Line 2 is 42 ÷ 5 + 42  

Hence there is a mistake in Line 2

3 0
3 years ago
Miyoko's goal is to earn more than \$950$950dollar sign, 950 this week. She earns \$250$250dollar sign, 250 for every day she wo
GrogVix [38]

$250 c+ $ 180 g > $ 950

<u>Step-by-step explanation:</u>

As a cryptographer (c), Miyoko earns per day = $ 250

As a geologist (g) , Miyoko earns per day = $ 180

So the equation comes to be $250 c+ $ 180 g = $ 950

The equation can be rewritten to find c as, (950-180 g) / 250

The equation can be rewritten to find g as, (950 - 250 c) / 180

Plugin different values of c and g in the above 2 equations, we can find that ,

To achieve the goal, Miyoko requires to be a geologist for 3 days and crpytographist for 2 days.

5 0
3 years ago
Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.​
vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}

\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}

4 0
2 years ago
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