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riadik2000 [5.3K]
3 years ago
7

Write 0.632 in two other forms

Mathematics
2 answers:
DedPeter [7]3 years ago
5 0
632
------ welcom peeps
1000
PtichkaEL [24]3 years ago
5 0
6.32*10^-1
And maybe 632/1000
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Ronald is preparing lemonade for his friends. To make 6 cups of lemonade, he needs 1 cup of lemon juice. To make 24 cups of lemo
timurjin [86]

Answer:1 cup=6 lemonade

X cups= 24 lemonade

Therefore;

x=24/6 × 1 = 4 cups.


Therefore four cups a


Step-by-step explanation:


5 0
4 years ago
Read 2 more answers
40 = 90^2/16 sinx cosx find x
lbvjy [14]

Answer:

<h2>x = 4.545</h2>

Step-by-step explanation:

Given the expression

40=\frac{90^2}{16}  sinxcosx\\\\Cross\ multiplying;\\\\16*40 = 90^2 sinxcosx\\\\640 =  90^2 sinxcosx\\\\\frac{640}{8100} = sinxcosx\\ from\ trig\ identity, sin2x = 2sinxcosx\\sinxcosx = sin2x/2\\

Hence, \ \frac{640}{8100} = \frac{sin2x}{2}  \\\\\frac{2*640}{8100} = sin2x\\ \\\frac{1280}{8100}=sin2x\\ \\0.158 = sin2x\\\\2x = sin^{-1} 0.158\\\\2x = 9.09\\x = 9.09/2\\x =4.545

5 0
3 years ago
Find the median data 46,22,46,49,45,39,40,42,41,39,40,36
ddd [48]

Answer:

median is 40.5

Step-by-step explanation:

22, 36, 39, 39, 40, 40, 41, 42, 45, 46, 46, 49

half way between numbers in numerical order

lands between 40 and 41

40.5

7 0
2 years ago
Read 2 more answers
If you pick a gumball at random, put it back, and then pick another gumball at random, what is the probability of picking a blue
AleksAgata [21]
The answer is B) 1/50

Probability of picking Blue-2/10 which is 1/5

Probability of picking Yellow-1/10

1/5x1/10=1/50
4 0
3 years ago
Read 2 more answers
Find the sum \[\sum_{k = 1}^{2004} \frac{1}{1 + \tan^2 (\frac{k \pi}{2 \cdot 2005})}.\]
vampirchik [111]

Answer:1002

Step-by-step explanation:

\Rightarrow \sum_{k=1}^{2004}\left ( \frac{1}{1+\tan ^2\left ( \frac{k\pi }{2\cdot 2005}\right )}\right )

and 1+\tan ^2\theta =\sec^2\theta

and \cos \theta =\frac{1}{\sec \theta }  

\Rightarrow \sum_{k=1}^{2004}\left ( \cos^2\frac{k\pi }{2\cdot 2005}\right )

as \cos ^2\theta =\cos ^2(\pi -\theta )

Applying this we get

\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]

every \thetathere exist \pi -\theta

such that \sin^2\theta +\cos^2\theta =1

therefore

\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]=1002                          

6 0
3 years ago
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